Bài tập phương trình và hệ phương trình - Diệp Tuân

Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 1 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 3 PHƯƠNG TRÌNH – HỆ PHƯƠNG TRÌNH A – LÝ THUYẾ T 1. Định nghĩa. Cho hai hàm số   y f x  và   y g x  có tập xác định lần lượt là f D và g D . Đặt fg D D D  . Mệnh đề chứa biến     "" f x g x  được gọi là phương trình một ẩn. x được gọi là ẩn số (hay ẩn) và D gọi là tập xác định của phương trình. 0 xD  gọi là một nghiệm của phương trình     f x g x  nếu     00 "" f x g x  là mệnh đề đúng. Chú ý: Các nghiệm của phương trình     f x g x  là các hoành độ giao điểm của hai đồ thị hàm số   y f x  và   y g x  . 2. Điều kiện của một phương trình Khi giải phương trình   1 , ta cần lưu ý với điều kiện đối với ẩn số x để   fx và   gx có nghĩa (tức là mọi phép toán đều thực hiện được). Ta cũng nói đó là điều kiện xác định của phương trình (hay gọi tắt là điều kiện của phương trình). 3. Phương trình nhiều ẩn Ngoài các phương trình một ẩn, ta còn gặp những phương trình có nhiều ẩn số, chẳng hạn     2 2 2 2 3 2 2 8, 2 4 2 3 2 . 3 x y x xy x xy z z xz y          Phương trình   2 là phương trình hai ẩn (x và y ), còn   3 là phương trình ba ẩn ( , xy và z ). Khi 2, 1 xy  thì hai vế của phương trình   2 có giá trị bằng nhau, ta nói cặp     ; 2;1 xy  là một nghiệm của phương trình   2. Tương tự, bộ ba số     ; ; 1;1;2 x y z  là một nghiệm của phương trình   3. 4. Phương trình chứa tham số Trong một phương trình (một hoặc nhiều ẩn), ngoài các chữ đóng vai trò ẩn số còn có thể có các chữ khác được xem như những hằng số và được gọi là tham số. 5. Phương trình tương đương, phương trình hệ quả. 5.1. Phương trình tương đương: Hai phương trình     11 f x g x  và     22 f x g x  được gọi là tương đương nếu chúng có cùng tập nghiệm. Kí hiệu là         1 1 2 2 f x g x f x g x    . Nhận xét: Phép biến đổi không làm thay đổi tập nghiệm của phương trình gọi là phép biến đổi tương đương.  Ví dụ 1. Tìm m để cặp phương trình sau tương đương. a).     2 1 0 1 x và     2 2 1 0 2 ax a x a     b).   2 9 0 1 x  và       2 2 5 3 1 0 2 x m x m      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... §B À I 1. Đ ẠI C ƯƠNG V Ề PH ƯƠNG TRÌNH Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 2 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 5.2. Phương trình hệ quả: Định nghĩa: Nếu mọi nghiệm của phương trình     11 f x g x  đều là nghiệm của phương trình     22 f x g x  thì phương trình     22 f x g x  được gọi là phương trình hệ quả của phương trình     11 . f x g x  Kí hiệu là         1 1 2 2 f x g x f x g x    Nhận xét: phương trình hệ quả có thể có thêm nghiệm không phải là nghiệm của phương trình ban đầu. Ta gọi đó là nghiệm ngoại lai.  Ví dụ 2. Giải các phương trình sau 2 15 1 36 x x x     Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 5.3. Các định lý: Định lý 1: Cho phương trình     f x g x  có tập xác định D ;   y h x  là hàm số xác định trên D . Khi đó trên D , phương trình đã cho tương đương với phương trình sau         f x h x g x h x            .. f x h x g x h x  nếu   0 hx  với mọi xD  Định lý 2: Khi bình phương hai vế của một phương trình, ta được phương trình hệ quả của phương trình đã cho.         22 f x g x f x g x    . Lưu ý: Khi giải phương trình ta cần chú ý Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 3 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Đặt điều kiện xác định(đkxđ) của phương trình và khi tìm được nghiệm của phương trình phải đối chiếu với điều kiện xác định. Nếu hai vế của phương trình luôn cùng dấu thì bình phương hai vế của nó ta thu được phương trình tương đương. Khi biến đổi phương trình thu được phương trình hệ quả thì khi tìm được nghiệm của phương trình hệ quả phải thử lại phương trình ban đầu để loại bỏ nghiệm ngoại lai. B. CÁC DẠNG TOÁN VÀ PHƯƠNG PHÁP GIẢI. DẠNG TOÁN 1: TÌM ĐIỀU KIỆN XÁC ĐỊNH CỦA PHƯƠNG TRÌNH. 1. Phương pháp. Điều kiện xác định của phương trình bao gồm các điều kiện để giá trị của     , f x g x cùng được xác định và các điều kiện khác (nếu có yêu cầu trong đề bài) Điều kiện để biểu thức    fx xác định là   0. fx     1 fx xác định là   0. fx     1 fx xác định là   0. fx  2. Bài tập minh họa.  Bài tập 1. Tìm điều kiện xác định của phương trình sau: a). 2 5 1 4 x x   b).1 3 2 xx     c). 1 2 3 3 2 xx     d). 3 1 42 32 x x xx    Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Tìm điều kiện xác định của phương trình sau rồi suy ra tập nghiệm của nó: a). 4 4 3 2 3 4 3 x x x      . b). 23 6 9 27 x x x      . c). 23 x x x      . d).     2 3 5 3 2 3 5 4 x x x x       . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 4 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Bài tập luyện tập.  Bài 1. Tìm điều kiện xác định của phương trình sau: a). 3 2 5 1 x xx   b).1 2 1 xx     c).1 2 4 2 4 xx     d). 2 1 26 32 x x xx    Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Tìm điều kiện xác định của phương trình sau rồi suy ra tập nghiệm của nó: a). 4 2 4 3 2 4 3 3 x x x      . b). 2 11 x x x      . c). 2 2 2 2 x x x      . d). 32 4 5 2 2 x x x x x       . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 5 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 4. Câu hỏi trắc nghiệm. Câu 1. Điều kiện xác định của phương trình 22 23 5 11 x xx   là A. 1. x  B. 1. x  C. 1. x  D. . x  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 2. Điều kiện xác định của phương trình 1 2 3 x x x      là A. 3. x  B. 2. x  C. 1. x  D. 3. x  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 3. Điều kiện xác định của phương trình 2 5 20 7 x x x      là A. 2. x  B. 7. x  C. 2 7. x  D. 2 7. x  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 4. Điều kiện xác định của phương trình 2 1 10 x x    là A. 0. x  B. 0. x  C. 0 x  và 2 1 0. x  D. 0 x  và 2 1 0. x  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 5. Điều kiện xác định của phương trình 2 8 22 x xx   là A. 2. x  B. 2. x  C. 2. x  D. 2. x  Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 6 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 6. Điều kiện xác định của phương trình 2 1 3 4 x x   là: A. 3 x  và 2. x  B. 2. x  C. 3 x  và 2. x  D. 3. x  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 7. Điều kiện xác định của phương trình 2 1 4 2 x x   là A. 2 x  hoặc 2. x  B. 2 x  hoặc 2. x  C. 2 x  hoặc 2. x  D. 2 x  hoặc 2. x  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 8. Điều kiện xác định của phương trình 1 3 2 24 x x x x    là A. 2 x  và 0. x  B. 2, 0 xx    và 3 . 2 x  C. 2 x  và 3 . 2 x  D. 2 x  và 0. x  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 9. Điều kiện xác định của phương trình 1 4 3 2 1 2 x x x x       là A. 2 x  và 1. x  B. 2 x  và 4 . 3 x  C.   4 2; \ 1 . 3 x        D. 2 x  và 1. x  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 10. Điều kiện xác định của phương trình 2 21 0 3 x xx    là A. 1 . 2 x  B. 1 2 x  và 3. x  C. 1 2 x  và 0. x  D. 0. x  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 7 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... DẠNG TOÁN 2: GIẢI PHƯƠNG TRÌNH BẰNG PHÉP BIẾN ĐỔI TƯƠNG ĐƯƠNG VÀ HỆ QUẢ 1. Phương pháp. Để giải phương trình ta thực hiện các phép biến đổi để đưa về phương trình tương đương với phương trình đã cho đơn giản hơn trong việc giải nó. Một số phép biến đổi thường sử dụng Cộng (hoặc trừ) cả hai vế của phương trình mà không làm thay đổi điều kiện xác định của phương trình ta thu được phương trình tương đương phương trình đã cho. Nhân (chia) vào hai vế với một biểu thức khác không và không làm thay đổi điều kiện xác định của phương trình ta thu được phương trình tương đương với phương trình đã cho. Bình phương hai vế của phương trình ta thu được phương trình hệ quả của phương trình đã cho. Sau đó ta thử lại nghiệm của phương trình. Bình phương hai vế của phương trình(hai vế luôn cùng dấu) ta luôn thu được phương trình tương đương với phương trình đã cho. 2. Bài tập minh họa.  Bài tập 3. Giải các phương trình sau a). 2 15 1 36 x x x     b). 2 1 2 22 x x xx     c). 42 3( 3 2) 0 x x x     d). 2 1( 2) 0 x x x     Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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.......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 4. Giải các phương trình sau Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 8 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 a). 2 2 3 4 15 xx    b). 2 3 4 8 3 x x x     . c). 2 1 2 xx    d). 2 1 1 xx    Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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.........................................................................................................................................................................................................  Bài tập 5. Tìm nghiệm   ; xy với x là số nguyên dương của phương trình sau 22 20 8 6 7 4 x x y y x      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 9 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 6. Tìm m để cặp phương trình sau tương đương a).   2 2 1 2 0 mx m x m      (1) và   22 2 3 15 0 m x x m      (2) b). 2 2 2 0 x mx    (3) và     32 2 4 2 1 4 0 x m x m x       (4) Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Bài tập tự luyện.  Bài 3. Giải các phương trình sau a). 2 16 1 24 xx   b). 21 3 33 x x xx     c). 2 1( 16) 0 xx    d). 2 3 0 23 x xx    Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 10 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài 4. Giải các phương trình sau a). 2 28 xx    b). 2 3 9 1 x x x     . c). 2 3 2 3 xx    d). 2 1 3 4 xx    Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Tìm m để cặp phương trình sau tương đương a). 3 2 0 x (1) và   3 4 0 m x m     (2) b). 20 x (1) và   22 3 2 2 0 m x x m x      (2) c). 2 10 x mx    (1) và     2 1 2 2 3 0 m x m x m       (2) d).     22 2 2 2 1 17 0 m x m x m m        (3) và   22 2 3 15 0 m x x m      (4) Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 11 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu hỏi trắc nghiệm. Câu 11. Hai phương trình được gọi là tương đương khi A. Có cùng dạng phương trình. B. Có cùng tập xác định. C. Có cùng tập hợp nghiệm. D. Cả A, B, C đều đúng. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 12. Phương trình nào sau đây tương đương với phương trình 2 40 x  ? A.     2 2 2 1 0. x x x      B.     2 2 3 2 0. x x x     C. 2 3 1. x  D. 2 4 4 0. xx    Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 12 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 13. Phương trình nào sau đây tương đương với phương trình 2 30 xx  ? A. 2 2 3 2. x x x x      B. 2 11 3. 33 xx xx     C. 2 3 3 3. x x x x    D. 2 2 2 1 3 1. x x x x      Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu 14. Cho phương trình       2 1 – 1 1 0 x x x    . Phương trình nào sau đây tương đương với phương trình đã cho ? A. 1 0. x B. 1 0. x C. 2 1 0. x  D.     – 1 1 0 . xx Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 13 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 15. Phương trình nào sau đây không tương đương với phương trình 1 1 x x  ? A. 2 1. xx    B. 2 1 2 1 0. xx     C. 5 0. xx D. 7 6 1 18. x     Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu 16. Khẳng định nào sau đây là đúng? A. 22 3 2 3 2. x x x x x x        B. 2 1 3 1 9 . x x x x      C. 22 3 2 2 3 . x x x x x x        D.   2 23 1 2 3 1 . 1 x x x x x         Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 17. Khẳng định nào sau đây là sai? Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 14 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 A. 1 2 1 1 0. x x x       B. 2 1 1 0 0. 1 x x x       C.     22 2 1 2 1 . x x x x        D. 2 1 1. xx    Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 18. Chọn cặp phương trình tương đương trong các cặp phương trình sau: A. 1 1 1 x x x      và 1. x  B. 2 1 2 x x x      và 1. x  C.   2 x x x  và 2 1. x D.   2 x x x  và 2 1. x Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Câu 19. Chọn cặp phương trình tương đương trong các cặp phương trình sau: A. 2 3 1 3 x x x      và 2 1. x  B. 1 0 1 xx x    và 0. x  C. 12 xx    và   2 1 2 . xx    D. 2 1 2 x x x      và 1. x  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 15 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu 20. Chọn cặp phương trình không tương đương trong các cặp phương trình sau: A. 2 12 x x x    và   2 2 1 . xx    B.3 1 8 3 x x x    và 6 1 16 3 . x x x    C. 22 32 x x x x x     và 3 2 . x x x  D. 22 xx  và 5 . 3 x  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 21. Tìm giá trị thực của tham số m để cặp phương trình sau tương đương: 2 2 2 0 x mx      1 và     32 2 4 2 1 4 0 x m x m x         2 . A. 2. m  B. 3. m  C. 1 . 2 m  D. 2. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 16 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 22. Tìm tất cả các giá trị thực của tham số m để cặp phương trình sau tương đương:   2 2 1 2 0 mx m x m        1 và   22 2 3 15 0 m x x m        2 . A. 5. m  B. 5; 4. mm    C. 4. m  D. 5. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu 23. Khẳng định nào sau đây là sai? A. 2 1 2 1. xx      B.   1 1 1. 1 xx x x      C. 2 3 2 3 8 4 5 0. x x x x        D. 3 9 2 3 12 0. x x x       Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 24. Cho phương trình 2 20 xx  . Trong các phương trình sau đây, phương trình nào không phải là hệ quả của phương trình đã cho? Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 17 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 A. 2 0. 1 x x x   B. 3 4 0. xx  C.     2 2 2 2 5 0. x x x     D. 32 2 0. x x x    Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Câu 25. Cho hai phương trình:       2 3 2 1 x x x    và     2 32 2 xx x    . Khẳng định nào sau đây là đúng? A. Phương trình   1 là hệ quả của phương trình   2 . B. Phương trình   1 và   2 là hai phương trình tương đương. C. Phương trình   2 là hệ quả của phương trình   1 . D. Cả A, B, C đều sai. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 18 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 26. Tập nghiệm của phương trình 22 22 x x x x    là: A.   0. S  B. . S  C.   0;2 . S  D.   2. S  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 27. Phương trình   2 1 1 0 x x x    có bao nhiêu nghiệm? A. 0. B. 1. C. 2. D. 3. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 28. Phương trình 23 6 9 27 x x x      có bao nhiêu nghiệm? A. 0. B. 1. C. 2. D. 3. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 29. Phương trình     2 3 5 3 2 3 5 4 x x x x       có bao nhiêu nghiệm? A. 0. B. 1. C. 2. D. 3. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 19 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Câu 30. Phương trình 11 x x x     có bao nhiêu nghiệm? A. 0. B. 1. C. 2. D. 3. 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Câu 31. Phương trình 2 2 2 2 x x x      có bao nhiêu nghiệm? A. 0. B. 1. C. 2. D. 3. 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Câu 32. Phương trình 32 4 5 2 2 x x x x x       có bao nhiêu nghiệm? A. 0. B. 1. C. 2. D. 3. 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Câu 33. Phương trình 1 2 1 11 x x xx    có bao nhiêu nghiệm? A. 0. B. 1. C. 2. D. 3. 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Câu 34. Phương trình   2 3 2 3 0 x x x     có bao nhiêu nghiệm? A. 0. B. 1. C. 2. D. 3. Lời giải. Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 20 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 35. Phương trình   2 2 1 0 x x x     có bao nhiêu nghiệm? A. 0. B. 1. C. 2. D. 3. 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A. LÝ THUYẾ T. I. Phương trình bậc nhất một ẩn. 1. Định nghĩa. Phương trình bậc nhất một ẩn là phương trình có dạng   0 1 ax b  với , ab  và 0 a  2. Giải và biện luận phương trình 0 ax b  (1). Nếu 0 a  :   1 b x a    do đó phương trình có nghiệm duy nhất . b x a  Nếu 0 a  : phương trình (1) trở thành 00 xb   Trường hợp 1: Với 0 b  phương trình nghiệm đúng với mọi xR  .  Trường hợp 2: Với 0 b  phương trình vô nghiệm. 3. Ví dụ minh họa .  Ví dụ 1: Giải và biện luận phương trình sau với m là tham số. a).   1 2 0 m x m     b).   1 9 3 m mx x    c). 2 ( 1) (3 7) 2 m x m x m      d).   2 1 2 1 2 mx m x    Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... §B À I 2. H ƯƠNG TRÌNH B ẬC NH ẤT VÀ B ẬC HAI M ỘT ẨN Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 21 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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II. Phương trình bậc hai một ẩn. 1. Định nghĩa: phương trình bậc hai một ẩn phương trình có dạng   2 2 0 ax bx c    với ,, abc là số thực và 0 a  2. Giải và biện luận phương trình 2 0 ax bx c    Nếu 0 a  : trở về giải và biện luận phương trình dạng (1) Nếu 0 a  : 2 4 b ac     Trường hợp 1: 0  phương trình có hai nghiệm phân biệt 2 b x a     .  Trường hợp 2: 0  phương trình có nghiệm kép 2 b x a   Trường hợp 3: 0  phương trình vô nghiệm. Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 22 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 3. Ví dụ minh họa.  Ví dụ 2: Giải và biện luận phương trình sau với m là tham số. a). 2 0 x x m    b).   2 1 2 2 0 m x mx m      c).   22 2 5 2 4 2 0 m m x mx      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 23 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... III. Định lí Vi-ét và ứng dụng. 1). Định lí Vi-ét. Hai số 1 x và 2 x là các nghiệm của phương trình bậc hai 2 0 ax bx c    khi và chỉ khi chúng thỏa mãn hệ thức 12 b xx a    và 12 c xx a  . 2) Ứng dụng. Nhẩm nghiệm của phương trình bậc hai. Phân tích thành nhân tử:  Nếu đa thức   2 f x ax bx c    có hai nghiệm 1 x và 2 x thì nó có thể phân tích thành nhân tử       12 f x a x x x x    . Tìm hai số khi biết tổng và tích của chúng:  Nếu hai số có tổng là S và tích là P thì chúng là nghiệm của phương trình 2 0 x Sx P    . Xét dấu của các nghiệm phương trình bậc hai: Cho phương trình bậc hai 2 0 ax bx c    (*), kí hiệu S, bc P aa    khi đó  Phương trình (*) có hai nghiệm trái dấu khi và chỉ khi 0 P   Phương trình (*) có hai nghiệm dương khi và chỉ khi 0 0 0 P S          Phương trình (*) có hai nghiệm âm khi và chỉ khi 0 0 0 P S         3. Ví dụ minh họa.  Ví dụ 3. Gọi 12 , xx là hai nghiệm của phương trình: 2 3 1 0 xx     . Tính giá trị của các biểu thức thỏa a). 22 12 A x x  ; b).     33 1 1 2 2 11 B x x x x     ; c). 22 12 11 C xx  . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Ví dụ 4. Tìm m để phương trình   22 3 4 1 4 1 0 x m x m m       có hai nghiệm phân biệt 12 , xx thỏa mãn:   12 12 1 1 1 2 xx xx    . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 24 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... B. CÁC DẠNG TOÁN VÀ PHƯƠNG PHÁP GIẢI. DẠNG TOÁN 1: GIẢI VÀ BIỆN LUẬN PHƯƠNG TRÌNH DẠNG 0 ax b  . 1. Phương pháp. Để giải và biện luận phương trình dạng 0 ax b  ta dựa vào kết quả sau Phương trình 0 ax b  có nghiệm 0 0 a ab       Phương trình 0 ax b  vô nghiệm 0 0 a b       Phương trình 0 ax b  có nghiệm duy nhất 0 a  2. Bài tập minh họa.  Bài tập 1. Giải và biện luận phương trình sau với , ab là tham số. a).     22 a x a b x b    b).     2 2 1 b ax b ax     Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 25 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 2. Tìm m để phương trình sau có nghiệm duy nhất. a). 22 ( ) 2 1 m m x x m     b).   4 3 2 ( 1) m mx m x m     Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 3. Tìm m để đồ thị hai hàm số sau không cắt nhau   22 13 y m x m x m     và   2 1 12 2 y m x x     . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 3. Bài tập luyên tập.  Bài 1. Giải và biện luận phương trình sau với m là tham số. a).   2 4 2 0 m x m     b). 2 ( 1) (3 1) 1 m x m x m      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 26 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Giải và biện luận các phương trình sau: a). 22 x a b x b a b a a b ab       (1) b).   2 2 1 12 1 1 1 ax ax x x x       (2) Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 27 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài 3. Tìm m để phương trình sau vô nghiệm. a). 22 ( ) 2 1 m m x x m     b).   2 32 m x m x m     Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài 4. Tìm điều kiện của , ab để phương trình sau có nghiệm . a).     2 1 1 a bx a a b x       b). 22 ( , 0) x a x b b a a b ab      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu hỏi trắc nghiệm. Câu 1. Tìm tất cả các giá trị thực của tham số m để phương trình   2 4 3 6 m x m    vô nghiệm. A. 1. m  B. 2. m  C. 2. m  D. 2. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 28 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 2. Tìm tất cả các giá trị thực của tham số m để phương trình 0 mx m  vô nghiệm. A. . m  B.   0. m  C. . m   D. . m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 3. Tìm giá trị thực của tham số m để phương trình   22 5 6 2 m m x m m     vô nghiệm. A. 1. m  B. 2. m  C. 3. m  D. 6. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 4. Cho phương trình     2 1 1 7 5 m x m x m      . Tìm tất cả các giá trị thực của tham số m để phương trình đã cho vô nghiệm. A. 1. m  B. 2; 3. mm  C. 2. m  D. 3. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 5. Cho hai hàm số   22 13 y m x m x m     và   2 1 12 2 y m x x     . Tìm tất cả các giá trị của tham số m để đồ thị hai hàm số đã cho không cắt nhau. A. 2. m  B. 2. m  C. 2. m  D. 1. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 6. Tìm tất cả các giá trị thực của tham số m để phương trình   2 4 2 m x m    có nghiệm duy nhất. A. 1. m  B. 2. m  C. 1. m  D. 2. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 29 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 7. Có bao nhiêu giá trị nguyên của tham số m thuộc đoạn   10;10  để phương trình     2 9 3 3 m x m m    có nghiệm duy nhất ? A. 2. B. 19. C. 20. D. 21. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 8. Gọi S là tập hợp tất cả các giá trị nguyên của tham số m thuộc đoạn   5;10  để phương trình     2 1 3 1 1 m x m x m      có nghiệm duy nhất. Tổng các phần tử trong S bằng: A. 15. B. 16. C. 39. D. 40. 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Câu 9. Tìm tất cả các giá trị thực của tham số m để phương trình   2 1 m m x m    có nghiệm duy nhất 1. x  A. 1. m  B. 0. m  C. 1. m  D. 1. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 10. Cho hai hàm số   2 12 y m x    và   37 y m x m    . Tìm tất cả các giá trị của tham số m để đồ thị hai hàm số đã cho cắt nhau. A. 2. m  B. 3. m  C. 2; 3. mm    D. 2; 3. mm    Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 11. Tìm tất cả các giá trị thực của tham số m để phương trình   2 11 m x m    có nghiệm đúng với mọi x thuộc . A. 1. m  B. 1. m  C. 1. m  D. 0. m  Lời giải. .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 30 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 12. Cho phương trình 2 6 4 3 . m x x m    Tìm tất cả các giá trị thực của tham số m để phương trình đã cho có nghiệm. A. 2. m  B. 2. m  C. 2 m  và 2. m  D. . m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 13. Cho phương trình   22 – 3 2 4 5 0 . m m x m m      Tìm tất cả các giá trị thực của tham số m để phương trình đã cho có nghiệm đúng với mọi x thuộc . A. 2. m  B. 5. m  C. 1. m  D. Không tồn tại. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 14. Cho phương trình   22 2 2. 3 m m x m m     Tìm tất cả các giá trị thực của tham số m để phương trình đã cho có nghiệm. A. 0. m  B. 2. m  C. 0; 2. mm  D. 0. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 15. Cho hai hàm số  11 y m x    và   2 31 y m x m    . Tìm tất cả các giá trị của tham số m để đồ thị hai hàm số đã cho trùng nhau. A. 2 1; . 3 mm    B. 1 m  và 2 . 3 m  C. 1. m  D. 2 . 3 m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... DẠNG TOÁN 2: GIẢI VÀ BIỆN LUẬN PHƯƠNG TRÌNH DẠNG 2 0 ax bx c    . Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 31 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 1. Phương pháp. Để giải và biện luận phương trình dạng 2 0 ax bx c    ta làm theo như các bước đã nêu ở trên. Nếu 0 a  : trở về giải và biện luận phương trình dạng (1) Nếu 0 a  : 2 4 b ac     Trường hợp 1: 0  phương trình có hai nghiệm phân biệt 2 b x a      Trường hợp 2: 0  phương trình có nghiệm kép 2 b x a   Trường hợp 3: 0  phương trình vô nghiệm. 2. Bài tập minh họa.  Bài tập 4. Giải và biện luận phương trình sau với , ab là tham số   2 2 2 0 ax a b x a b      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 5. Tìm m để phương trình 2 10 mx x m     a). Có nghiệm kép. b). Có hai nghiệm phân biệt. Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 32 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 3. Bài tập luyện tập  Bài 5. Tìm m để phương trình 22 3 (2 1) 0 x mx m m      có nghiệm kép tìm nghiệm kép đó Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài 6. Cho phương trình: 2 2 1 0 mx mx m     a). Giải phương trình đã cho khi 2 m  . b). Tìm m để phương trình đã cho có nghiệm Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài 7. Giải và biện luận phương trình a). 2 ( 2) 2( 1) 5 0 m x m x m       b). 2 ( 2) (2 1) 2 0 m x m x m       Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 33 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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.......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài 8. Tùy thuộc vào giá trị của tham số m , hãy tìm hoành độ giao điểm của đường thẳng : 2 d y x m  và Parabol (P):   2 – 1 2 3 – 1 . y m x mx m    Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 4. Câu hỏi trắc nghiệm Câu 16. Cho hai hàm số  11 y m x    và   2 31 y m x m    . Tìm tất cả các giá trị của tham số m để đồ thị hai hàm số đã cho trùng nhau. A. 2 1; . 3 mm    B. 1 m  và 2 . 3 m  C. 1. m  D. 2 . 3 m  Lời giải. .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 34 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 17. Phương trình 2 0 ax bx c    có nghiệm duy nhất khi và chỉ khi: A. 0. a  B. 0 0 a      hoặc 0 . 0 a b      C. 0. abc    D. 0 . 0 a      Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 18. Số 1  là nghiệm của phương trình nào trong các phương trình sau? A. 2 4 2 0. xx    B. 2 2 5 7 0. xx    C. 2 3 5 2 0. xx     D. 3 1 0. x  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 19. Nghiệm của phương trình 2 7 12 0 xx    có thể xem là hoành độ giao điểm của hai đồ thị hàm số nào sau đây? A. 2 yx  và 7 12. yx    B. 2 yx  và 7 12. yx    C. 2 yx  và 7 12. yx  D. 2 yx  và 7 12. yx  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 20. Có bao nhiêu giá trị nguyên của tham số thực m thuộc đoạn   10;10  để phương trình 2 0 x x m    vô nghiệm? A. 9. B. 10. C. 20. D. 21. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 35 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 21. Phương trình   2 1 2 2 0 m x mx m      vô nghiệm khi: A. 2. m  B. 2. m  C. 2. m  D. 2. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 22. Số nguyên k nhỏ nhất thỏa mãn phương trình   2 2 4 6 0 x kx x     vô nghiệm là? A. 1. k  B. 1. k  C. 2. k  D. 3. k  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 23. Phương trình   2 – 2 2 – 1 0 m x x  có nghiệm kép khi: A. 1; 2. mm  B. 1. m  C. 2. m  D. 1. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 24. Phương trình 2 6 4 3 mx x m    có nghiệm duy nhất khi: A. . m  B. 0. m  C. . m  D. 0. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 36 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 25. Phương trình   2 – 2 1 1 0 mx m x m     có nghiệm duy nhất khi: A. 0. m  B. 1. m  C. 0; 1. mm    D. 1. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 26. Phương trình     2 1 – 6 1 2 3 0 m x m x m      có nghiệm kép khi: A. 1. m  B. 6 1; 7 mm     C. 6 . 7 m  D. 6 . 7 m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 27. Phương trình     2 2 1 1 x x mx    có nghiệm duy nhất khi: A. 17 . 8 m  B. 2. m  C. 17 2; . 8 mm  D. 1. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 28. Gọi S là tập hợp tất cả các giá trị thực của tham số m để phương trình   2 2 2 1 2 0 m x x m      có nghiệm duy nhất. Tổng của các phần tử trong S bằng: A. 5 . 2 B. 3. C. 7 . 2 D. 9 . 2 Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 37 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 29. Phương trình   2 1 6 1 0 m x x     có hai nghiệm phân biệt khi: A. 8. m  B. 5 . 4 m  C. 8; 1. mm    D. 5 ; 1. 4 mm    Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 30. Có bao nhiêu giá trị nguyên của tham số thực m thuộc đoạn   5;5  để phương trình   2 2 2 1 0 mx m x m      có hai nghiệm phân biệt. A. 5. B. 6. C. 9. D. 10. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 31. Phương trình     22 2 2 3 0 m x m x      có hai nghiệm phân biệt khi: A. 0 2. m  B. 2. m  C. . m  D. 2. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 32. Tìm giá trị thực của tham số m để đường thẳng :2 d y x m  tiếp xúc với parabol     2 – 1 2 3 – 1 : . P y m x mx m    A. 1. m  B. 1. m  C. 0. m  D. 2. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 38 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 33. Phương trình 2 0 xm  có nghiệm khi: A. 0. m  B. 0. m  C. 0. m  D. 0. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 34. Gọi S là tập hợp tất cả các giá trị nguyên của tham số m thuộc   20;20  để phương trình 2 2 144 0 x mx    có nghiệm. Tổng của các phần tử trong S bằng: A. 21. B. 18. C. 1. D. 0. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 35. Tìm tất cả các giá trị thực của tham số m để hai đồ thị hàm số 2 23 y x x     và 2 y x m  có điểm chung. A. 7 . 2 m  B. 7 . 2 m  C. 7 . 2 m  D. 7 . 2 m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 36. Phương trình   2 1 3 1 0 m x x     có nghiệm khi: A. 5 . 4 m  B. 5 . 4 m  C. 5 . 4 m  D. 5 . 4 m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 39 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 37. Có bao nhiêu giá trị nguyên của tham số m thuộc đoạn   10;10  để phương trình 2 10 mx mx    có nghiệm. A. 17. B. 18. C. 20. D. 21. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 38. Biết rằng phương trình 2 4 1 0 x x m     có một nghiệm bằng 3 . Nghiệm còn lại của phương trình bằng: A. 1.  B. 1. C. 2. D. 4. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 39. Tìm tất cả các giá trị thực của tham số m để phương trình   2 3 2 1 0 x m x m      có một nghiệm gấp đôi nghiệm còn lại. A. 5 ;7 . 2 m     B. 1 2; . 2 m       C. 2 0; . 5 m     D. 3 ;1 . 4 m     Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 40 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 40. Tìm tất cả các giá trị thực của tham số m để phương trình   2 3 2 1 3 5 0 x m x m      có một nghiệm gấp ba nghiệm còn lại. A. 7. m  B. 3. m  C. 3; 7. mm  D. . m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu 41. Tìm tất cả các giá trị thực của tham số m để phương trình     2 1 4 4 0 x x mx     ba nghiệm phân biệt. A. . m  B. 0. m  C. 3 . 4 m  D. 3 . 4 m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... DẠNG TOÁN 3: MỘT SỐ ỨNG DỤNG CỦA ĐỊNH LÝ VI -ÉT. Loại 1: Nhẩm nghiệm phương trình bậc hai, phân tích thành nhân tử. Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 41 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880  Bài tập 6. Cho phương trình 2 2 5 0 x mx    . Biết phương trình có một nghiệm là 2. Tìm m và tìm nghiệm còn lại Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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.......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 7. Phân tích đa thức sau thành nhân tử a). 2 ( ) 3 14 8 f x x x    b). 42 ( ) 5 4 g x x x     c). 22 ( ; ) 6 11 3 P x y x xy y    . d). 22 ( ; ) 2 2 3 2 Q x y x y xy x y      . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 42 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 8. Phân tích đa thức   4 2 2 2 f x x mx x m m      thành tích của hai tam thức bậc hai ẩn x . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Loại 2: Bài toán liên quan đến biểu thức đối xứng hai nghiệm 12 , xx của phương trình bậc hai.  Bài tập 9. Cho phương trình   22 2 1 2 0 x m x m      với m là tham số. Tìm m để phương trình có hai nghiệm 12 ; xx sao cho a).   33 1 2 1 2 1 2 2 x x xx x x    b). 4 4 2 12 16 64 x x m m    c).   1 2 1 2 26 A xx x x     đạt giá trị nhỏ nhất d).   22 1 2 1 2 2 16 3 B x x xx     đạt giá trị lớn nhất Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 43 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Cho phương trình 2 10 x mx m    với m là tham số. a). Chứng minh rằng phương trình luôn có nghiệm với mọi m b). Gọi 12 , xx là hai nghiệm của phương trình. Tìm hệ thức liên hệ giữa 12 , xx không phụ thuộc vào m c). Tìm giá trị nhỏ nhất và lớn nhất của biểu thức 12 22 1 2 1 2 23 2( 1) xx A x x xx      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 44 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Chú ý: Để tìm giá trị lớn nhất, giá trị nhỏ nhất của biểu thức 2 21 2 m A m    ta làm như sau Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 3. Bài tập luyện tập.  Bài 9. Phân tích đa thức sau thành nhân tử a). 2 ( ) 2 5 3 f x x x    b). 42 ( ) 2 14 36 g x x x    c). 22 ( ; ) 3 5 2 P x y x xy y    . d). 22 ( ; ) 2 3 1 Q x y x y xy y      . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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.......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 10. Phân tích đa thức     3 2 2 2 1 2 f x m x x x m m m     (biến x với tham số m ) thành tích một đã thức bậc hai và một bậc nhất. Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 45 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài11. Cho phương trình   22 2 1 3 0 x m x m      với m là tham số. Tìm m để phương trình có hai nghiệm 12 ; xx sao cho a). 1 2 1 2 2 x x x x  b).   22 1 2 1 2 2 A x x xx    đạt giá trị lớn nhất c). 12 22 1 2 1 2 xx B x x x x   đạt giá trị nhỏ nhất Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Câu hỏi trắc nghiệm. Câu 42. Phương trình   2 0 0 ax bx c a     có hai nghiệm phân biệt cùng dấu khi và chỉ khi: A. 0 . 0 P      B. 0 . 0 P      C. 0 . 0 S      D. 0 . 0 S      Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 46 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 43. Phương trình   2 0 0 ax bx c a     có hai nghiệm âm phân biệt khi và chỉ khi: A. 0 . 0 P      B. 0 0. 0 P S         C. 0 0. 0 P S         D. 0 . 0 S      Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 44. Phương trình   2 0 0 ax bx c a     có hai nghiệm dương phân biệt khi và chỉ khi: A. 0 . 0 P      B. 0 0. 0 P S         C. 0 0. 0 P S         D. 0 . 0 S      Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 45. Phương trình   2 0 0 ax bx c a     có hai nghiệm trái dấu khi và chỉ khi: A. 0 . 0 S      B. 0 . 0 S      C. 0. P  D. 0. P  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 47 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 46. Phương trình 2 10 x mx    có hai nghiệm âm phân biệt khi: A. 2. m  B. 2. m  C. 2. m  D. 0. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 47. Có bao nhiêu giá trị nguyên của tham số m thuộc   5;5  để phương trình 22 40 x mx m    có hai nghiệm âm phân biệt? A. 5. B. 6. C. 10. D. 11. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 48. Tập hợp tất cả các giá trị thực của tham số m để phương trình 2 0 mx x m    có hai nghiệm âm phân biệt là: A. 1 ;0 . 2 m     B. 11 ;. 22 m     C.   0;2 . m  D. 1 0; . 2 m     Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 49. Gọi S là tập tất cả các giá trị nguyên của tham số m thuộc đoạn   2;6  để phương trình 22 40 x mx m    có hai nghiệm dương phân biệt. Tổng các phần tử trong S bằng: A. 3.  B. 2. C. 18. D. 21. Lời giải. .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 48 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 50. Tập hợp tất cả các giá trị thực của tham số m để phương trình   22 2 1 1 0 x m x m      có hai nghiệm dương phân biệt là: A.   1 ;1 . m B.   1 ; . m    C. 1 ;. 2 m        D.   ; 1 . m     Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 51. Phương trình   2 1 3 1 0 m x x     có hai nghiệm trái dấu khi: A. 1. m  B. 1. m  C. 1. m  D. 1. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 52. Giả sử phương trình   22 2 1 2 0 x m x m      (m là tham số) có hai nghiệm là 12 , xx . Tính giá trị biểu thức   1 2 1 2 35 P xx x x    theo . m A. 2 3 10 6. P m m    B. 2 3 10 5. P m m    C. 2 3 10 1. P m m    D. 2 3 10 1. P m m    Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 49 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Câu 53. Giả sử phương trình 2 30 x x m    (m là tham số) có hai nghiệm là 12 , xx . Tính giá trị biểu thức     22 1 2 2 1 11 P x x x x     theo . m A. 9. Pm    B. 5 9. Pm  C. 9. Pm  D. 5 9. Pm    Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu 54. Giả sử phương trình 2 2 4 1 0 x ax    có hai nghiệm 12 , . xx Tính giá trị của biểu thức 12 . T x x  A. 2 42 . 3 a T   B. 2 4 2. Ta  C. 2 8 . 2 a T   D. 2 8 . 4 a T   Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 55. Cho phương trình 2 0 x px q    trong đó 0, 0. pq  Nếu hiệu các nghiệm của phương trình bằng 1. Khi đó p bằng A. 4 1. q  B. 4 1. q  C. 4 1. q  D. 1. q  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 50 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 56. Gọi 12 , xx là hai nghiệm của phương trình   22 2 1 1 0 x m x m      (m là tham số). Tìm giá trị nguyên của m sao cho biểu thức 12 12 xx P xx   có giá trị nguyên. A. 2. m  B. 1. m  C. 1. m  D. 2. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu 57. Gọi 12 , xx là hai nghiệm của phương trình   22 2 1 2 0 x m x m      (m là tham số). Tìm m để biểu thức   1 2 1 2 26 P xx x x     đạt giá trị nhỏ nhất. A. 1 . 2 m  B. 1. m  C. 2. m  D. 12. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 51 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 58. Gọi 12 , xx là hai nghiệm của phương trình 22 2 2 2 0 x mx m     (m là tham số). Tìm giá trị lớn nhất max P của biểu thức 1 2 1 2 2 4 . P xx x x     A. 2 0 x ax b    B. max 2. P  C. max 25 . 4 P  D. max 9 . 4 P  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 59. Gọi 12 , xx là hai nghiệm của phương trình   22 2 1 2 3 1 0 x m x m m       (m là tham số). Tìm giá trị lớn nhất max P của biểu thức 1 2 1 2 . P x x xx    A. max 1 . 4 P  B. max 1. P  C. max 9 . 8 P  D. max 9 . 16 P  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 60. Gọi 12 , xx là hai nghiệm của phương trình 2 10 x mx m    (m là tham số). Tìm m để biểu thức   12 22 1 2 1 2 23 21 xx P x x xx      đạt giá trị lớn nhất. Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 52 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 A. 1 . 2 m  B. 1. m  C. 2. m  D. 5 . 2 m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu 61. Gọi 12 , xx là hai nghiệm của phương trình 2 10 x mx m    (m là tham số). Tìm giá trị nhỏ nhất min P của biểu thức   12 22 1 2 1 2 23 . 21 xx P x x xx      A. min 2. P  B. min 1 . 2 P  C. min 0. P  D. min 1. P  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... DẠNG TOÁN 4: Một số bài toán liên quan đến nghiệm của phương trình bậc hai. Bài toán 1: Tìm điều kiện để hai phương trình bậc hai 2 0 ax bx c    và 2 0 ax bx c       có nghiệm chung. Chúng ta làm như sau: 1. Phương pháp. Bước 1: Giả sử hai phương trình có nghiệm chung là 0 x thì 2 00 / 2 / / 00 0 0 ax bx c a x b x c            Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 53 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Giải hệ tìm được 0 x ,suy ra giá trị của tham số Bước 2: Thế giá trị của tham số tìm được vào hai phương trình để kiểm tra và kết luận. 2. Bài tập minh họa.  Bài tập 11. Tìm tất cả các giá trị của a để hai phương trình 2 10 x ax    và 2 0 x x a    có nghiệm chung Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 12. Tìm tất cả các giá trị của m để phương trình 22 ( 2 1)( ) 2 30 x mx m x m x       có bốn nghiệm phân biệt Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 54 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài toán 2: Chứng minh trong các phương trình bậc hai có ít nhất một phương trình có nghiệm 1. Phương pháp. Để giải quyết bài toán này chúng ta sẽ đi chứng minh tổng các biệt thức Delta là một số không âm. 2. Bài tập minh họa.  Bài tập 13. Cho các số dương ,, abc thỏa mãn diệu kiện 2 3 1 a b c    .Chứng minh rằng có ít nhất một trong hai phương trình sau có nghiệm     22 22 4 4 2 1 4 192 1 0 4 4 2 1 4 96 1 0 x a x a abc x b x b abc             Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 14. Cho các số ,, abc thỏa mãn điệu kiện 6 abc    .Chứng minh rằng có ít nhất một trong ba phương trình sau có nghiệm 2 2 2 10 10 10 x ax x bx x cx          Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 55 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài toán 3: Chứng minh bất đẳng thức có chứa các hệ số của phương trình bậc hai với nghiệm của nó có điều kiện. 1. Phương pháp. Để làm xuất hiện điều kiện ràng buộc đối với hệ số phương trình bậc hai ta thường dựa trên Nếu phương trình bậc hai 2 0 ax bx c    có nghiệm thực thì 2 04 b ac     . Sử dụng định lí Viét và điều kiện nghiệm của đề bài đã cho để suy ra ràng buộc của hệ số ,, abc 2. Bài tập minh họa.  Bài tập 15.Cho phương trình 2 0 x bx c    có hai nghiệm thực dương 12 , xx thoả mãn 12 1. xx  Chứng minh rằng: a). 1 . 4 c  b). ( 1) 5 . b c c  Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 16.Cho phương trình 2 0 x bx c    có hai nghiệm thực dương 12 , xx thoả mãn 12 1. xx  a). Chứng minh rằng: 2 1 2. 2 bc  b). Tìm giá trị lớn nhất của biểu thức: 3 2 3 1 P bc b b     . Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 3. Bài tập luyện tập. Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 56 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Bài 12. Tìm m để hai phương trình sau có nghiệm chung 2 2 4 1 0 x mx m     (1) và   2 3 1 2 1 0 x m x m      (2). Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài 13. Chứng minh rằng nếu hai phương trình 2 0 x ax b    và 2 0 x mx n    có nghiệm chung thì       2 n b m a an bm     . Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài 14. Cho ,, abc là các số thực không đồng thời bằng 0. Chứng minh rằng trong ba phương trình sau có ít nhất một phương trình có nghiệm 2 20 ax bx c    (1); 2 20 bx cx a    (2); 2 20 cx bx b    (3). Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 57 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 15.Cho phương trình 2 0 x bx c    có hai nghiệm thực dương 12 , xx thoả mãn 12 1. xx  a). Chứng minh rằng: 2 4. b  b). Tìm giá trị nhỏ nhất của biểu thức: 2 2 3 4 2 . 1 b c b P b      Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài 16. Giả sử phương trình bậc hai 2 0 ax bx c    có hai nghiệm thuộc [0;3] . Tìm giá trị lớn nhất và giá trị nhỏ nhất của biểu thức: 22 2 18 9 93 a ab b Q a ab ac    Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 58 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 17. Cho phương trình bậc hai 2 0 ax x c    có hai nghiệm thực dương 12 , xx thoả mãn 12 1. xx  Tìm giá trị nhỏ nhất của biểu thức 2 23 . ac P a c a    Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 4. Câu hỏi trắc nghiệm. Câu 62. Nếu 0 m  và 0 n  là các nghiệm của phương trình 2 0 x mx n    thì tổng mn  bằng: A. 1 . 2  B. 1.  C. 1 . 2 D. 1. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 63. Giả sử các nghiệm của phương trình 2 0 x px q    là lập phương các nghiệm của phương trình 2 0 x mx n    . Mệnh đề nào sau đây đúng? A. 3 . p q m  B. 3 3. p m mn  C. 3 3. p m mn  D. 3 . mp nq     Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 59 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Câu 64. Cho hai phương trình 2 2 1 0 x mx    và 2 2 0. x x m    Có hai giá trị của m để phương trình này có một nghiệm là nghịch đảo của một nghiệm của phương trình kia. Tính tổng S của hai giá trị m đó. A. 5 . 4 S  B. 1. S  C. 1 . 4 S  D. 1 . 4 S  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 65. Cho hai phương trình 2 20 x mx    và 2 20 x x m    . Có bao nhiêu giá trị của m để một nghiệm của phương trình này và một nghiệm của phương trình kia có tổng là 3 ? A. 0. B. 1. C. 2. D. 3. 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Câu 66. Cho , , , a b c d là các số thực khác 0 . Biết c và d là hai nghiệm của phương trình 2 0 x ax b    và , ab là hai nghiệm của phương trình 2 0. x cx d    Tính giá trị của biểu thức . S a b c d     A. 2. S  B. 0. S  C. 15 . 2 S   D. 2. S  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 60 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... A–LÝ THUYẾ T 1. Phương trình chứa ẩn trong dấu giá trị tuyệt đối. Để giải phương trình chứa ẩn trong dấu giá trị tuyệt đối ta dùng các cách sau khử dấu giá trị tuyệt đối.  Định nghĩa của giá trị tuyệt đối ( ) ( ) ( ) 0 ( ) ( ) ( ) ( ) ( ) 0 f x g x fx f x g x f x g x fx                   Hoặc bình phương hai vế 22 ( ) 0 ( ) ( ) ( ) ( ) gx f x g x f x g x        Hoặc dùng tính chất của GTTĐ ( ) 0 ( ) ( ) ( ) ( ) ( ) ( ) gx f x g x f x g x f x g x             Ví dụ 1. Giải phương trình 3 2 1. xx      3 Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 2. Phương trình chứa ẩn dưới dấu căn Để giải các phương trình chứa ẩn dưới dấu căn bậc hai, ta thường bình phương hai vế để đưa về một phương trình hệ quả không chứa ẩn dưới dấu căn. §B À I 3. M ỘT S Ố PH ƯƠNG TRÌNH QUY V Ề B ẬC NH ẤT VÀ B ẬC HAI Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 61 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880   2 ( ) ( ) ( ) ( ) ( ) 0 f x g x f x g x gx         Ví dụ 2. Giải phương trình 2 3 2. xx    4 Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... B-PHÂN DẠNG VÀ BÀI TẬP MINH HỌA. DẠNG TOÁN 1. Phương tình chứa ẩn trong giá trị tuyệt đối. 1. Phương pháp Để giải phương trình chứa ẩn trong dấu giá trị tuyệt đối(GTTĐ) ta tìm cách để khử dấu GTTĐ, bằng cách: Phương pháp 1. Dùng định nghĩa của GTTĐ ( ) ( ) ( ) 0 ( ) ( ) ( ) ( ) ( ) 0 f x g x fx f x g x f x g x fx                  Phương pháp 2. Dùng tính chất của GTTĐ ( ) 0 ( ) ( ) ( ) ( ) ( ) ( ) gx f x g x f x g x f x g x             Nhận xét:  Nếu đa thức     , f x g x có bậc quá lớn (từ bậc hai trở lên) thì ta dùng định nghĩa hoặc tính chất.  Nếu phương trình không có dạng cơ bản. Phương pháp 3. Bình phương hai vế 22 ( ) 0 ( ) ( ) ( ) ( ) gx f x g x f x g x       Nhận xét:  Nếu đa thức     , f x g x có bậc một thì ta dùng phương pháp bình phương hai vế Phương pháp 4. Đặt ẩn phụ. Nhận xét:  Có nhiều đa thức chứa trị tuyệt đối giống nhau thì ta dùng phương pháp đặt ẩn phụ.  Đặt biệt phương trình dạng ( ) ( ) f x g x  ta có thể giải bằng cách biến đổi tương đương như sau ( ) ( ) ( ) ( ) ( ) ( ) f x g x f x g x f x g x       hoặc 22 ( ) ( ) ( ) ( ) f x g x f x g x     Nếu phương trình chứa nhiều trị tuyệt đối     . f x g x c  thì ta lập bảng xét dấu, xét nhiều khoảng. Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 62 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 2. Bài tập minh họa. Bài tập1. Giải các phương trình sau a). 2 2 1 3 4 x x x     . b). 3 2 3 2 xx    c). 2 4 5 4 17 x x x     d). 2 2 5 2 7 5 0 x x x      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 2: Giải các phương trình sau. a).   2 1 3 1 2 0 xx      b).   4 1 2 1 1 x x x     c).   2 2 2 9 2 2 1 2 7 1 1 xx xx x x        d).   2 2 4 3 x x x      Lời giải .......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 63 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 3. Giải và biện luận các phương trình sau a). 21 mx m mx x     (*) b). 2 1 1 mx x x     (**) Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 64 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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.........................................................................................................................................................................................................  Bài tập 4. Tìm m để phương trình 22 ( 1) 2 1 x x mx m x m       có ba nghiệm phân biệt. Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 65 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 3. Bài tập luyện tập. Bài 1. Giải các phương trình sau a). 2 | 3 2 | 2 3 x x x     b). 32 1 3 2 x x x     c). 2 11 xx    d). 2 1 1 1 1 x x x       e).   3 2 5 2 3 2 x x x x       . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 2. Giải các phương trình sau a).   2 2 1 3 2 1 4 0 xx      b). 4 2 2 2 6 4 2 x x x xx     Lời giải Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 66 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... Bài 3. Cho phương trình 2 2 2 1 3 0 x x x m       a). Giải phương trình khi 2 m  b). Tìm m để phương trình sau có nghiệm Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 4. Giải và biện luận các phương trình sau a). 21 mx m x    b). 21 mx x mx    Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 67 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 4. Câu hỏi trắc nghiệm. Câu 1. Tập nghiệm S của phương trình 3 2 3 2 xx    là: A.   1;1 . S B.   1. S C.   1. S  D. S    0. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 2. Phương trình 2 4 2 4 0 xx     có bao nhiêu nghiệm? A.0 . B.1. C. 2 . D. Vô số. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 3. Tập nghiệm S của phương trình 2 1 3 xx    là: A. 4 . 3 S     B. . S  C. 4 2; . 3 S     D.   2. S Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 4. Tổng các nghiệm của phương trình 2 5 4 4 x x x     bằng: A. 12.  B. 6.  C. 6. D. 12. Lời giải. Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 68 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 5. Gọi   1 2 1 2 , x x x x  là hai nghiệm của phương trình 2 4 5 4 17 x x x     . Tính giá trị biểu thức 2 12 . P x x  A. 16. P  B. 58. P  C. 28. P  D. 22. P  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 6. Tập nghiệm S của phương trình 2 3 5 xx    là: A. 37 ;. 24 S     B. 37 ;. 24 S     C. 73 ;. 42 S       D. 73 ;. 42 S     Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 7. Tổng các nghiệm của phương trình 2 2 2 xx    bằng: A. 1 . 2 B. 2 . 3 C. 6. D. 20 . 3 Lời giải. Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 69 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 8. Phương trình 2 2 1 3 4 x x x     có bao nhiêu nghiệm? A. 0 . B. 1 . C. 2 . D. 4 . Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 9. Phương trình 2 4 1 0 xx     có bao nhiêu nghiệm ? A. 0 . B. 1 . C. 2 . D. Vô số. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 10. Tổng các nghiệm của phương trình 2 2 5 2 7 5 0 x x x      bằng: A. 6. B. 5 . 2 C. 7 . 2 D. 3 . 2 Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 11. Phương trình   2 1 3 1 2 0 xx      có bao nhiêu nghiệm? A. 0. B. 1. C. 2. D. 4. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 70 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 12. Tổng các nghiệm của phương trình   4 1 2 1 1 x x x     bằng: A. 0. B. 1. C. 2. D. 2.  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 13. Với giá trị nào của a thì phương trình 3 2 1 x ax    có nghiệm duy nhất? A. 3 . 2 a  B. 3 . 2 a   C. 33 . 22 aa     D. 33 . 22 aa     Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu 14. Tìm giá trị thực của tham số m để phương trình 2 1 x x m    có nghiệm duy nhất. A. 0. m  B. 1. m  C. 1. m  D.Không có . m Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 71 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 15. Có bao nhiêu giá trị nguyên của tham số m thuộc đoạn   5;5  để phương trình 2 1 1 mx x x     có đúng hai nghiệm phân biệt? A. 8. B. 9. C. 10. D. 11. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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DẠNG TOÁN 2: PHƯƠNG TRÌNH CHỨA ẨN Ở MẪU 1. Phương pháp. Để giải phương trình chứa ẩn ở mẫu ta thường Quy đồng mẫu số (chú ý cần đặt điều kiện mẫu số khác không). Đặt ẩn phụ. 2. Bài tập minh họa.  Bài tập 5. Giải các phương trình sau a). 2 1 1 3 2 2 xx xx    b). 2 10 50 1 2 3 (2 )( 3) x x x x        . c). 22 3 4 2 ( 1) (2 1) xx xx    . d). 1 1 2 1 2 2 1 x x x x x x        Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 72 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 6. Giải các phương trình sau a). 4 3 2 1 2 1 2 2 2 3 2 4 x x x x        . b). 2 2 2 1 1 1 3 5 4 11 28 17 70 4 2 x x x x x x x           c).   2 2 45 1 2 x x   Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 73 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài tập 7. Giải và biện luận phương trình sau với m là tham số. a). 2 1 xm x    (1) b). 2 2 2 1 1 x mx x    (2) c). 2 2 26 3 x mx m x    (3) d). 32 1 x mx m x    (4) Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 74 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 75 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880  Bài tập 8. Tìm điều kiện của tham số a và b để phương trình   22 2 a b a b x a x b x a b x ab        (*) a). Có nghiệm duy nhất. b). Có nghiệm Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Bài tập luyện tập. Bài 5. Giải các phương trình sau: a). 22 13 1 6 2 21 2 7 9 x x x x      b). 3 2 2 2 4 1 1 4 2 3 8 12 2 3 4 2 7 6 x x x x x x x           c). 1 2 3 4 4 1 2 3 4 x x x x x x x x             Lời giải .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 76 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 6. Giải phương trình a). 22 2 13 6 3 5 2 3 2 xx x x x x      b). 42 32 31 3 xx x x x    c). 22 11 15 ( 1) xx   Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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.......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 7. Giải phương trình a). 22 1 1 2 12 2 3 3 x x x x x x                    b). 22 2( 1) 13( 1) 6 3 3 7 6 xx x x x x      Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 77 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... Bài 8. Giải và biện luận phương trình sau   2 2 1 12 1 1 1 ax ax x x x       Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 9. Tìm điều kiện , ab để phương trình 2 ab x b x a   có hai nghiệm phân biệt. Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 78 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 4. Câu hỏi trắc nghiệm. Câu 16. Tập nghiệm S của phương trình 33 2 11 x x xx   là: A. 3 1; . 2 S     B.   1. S  C. 3 . 2 S     D.   \ 1 . S  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 17. Tập nghiệm của phương trình 2 54 22 xx xx    là: A.   1;4 . S  B.   1. S  C. . S  D.   4. S  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 18. Phương trình 2 2 2 10 3 5 xx x xx    có bao nhiêu nghiệm? A. 0. B. 1. C. 2. D. 3. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 19. Gọi 0 x là nghiệm của phương trình     2 10 50 1 2 3 2 3 x x x x        . Mệnh đề nào sau đây đúng? A.   0 5; 3 . x    B.   0 3; 1 . x    C.   0 1;4 . x  D.   0 4; . x    Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 79 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Câu 20. Tập nghiệm S của phương trình   2 11 1 1 mx x    trong trường hợp 0 m  là: A. 2 1 . m S m      B. . S  C. . S  D. 2 2 . S m     Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 21. Tập nghiệm S của phương trình   2 2 3 6 3 m x m x   khi 0 m  là: A. . S  B. 3 . S m     C. . S  D.   \ 0 . S  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 22. Có bao nhiêu giá trị của tham số m để phương trình 2 2 2 1 1 x mx x    vô nghiệm? A. 0. B. 1. C. 2. D. 3. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 23. Phương trình 21 3 1 mx x    có nghiệm duy nhất khi: A. 3 . 2 m  B. 0. m  C. 0 m  và 3 . 2 m  D. 1 2 m  và 3 . 2 m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 80 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Câu 24. Gọi S là tập hợp các giá trị nguyên của tham số m thuộc đoạn   3;5  để phương trình 2 11 x m x xx    có nghiệm. Tổng các phần tử trong tập S bằng: A. 1.  B. 8. C. 9. D. 10. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 25. Có bao nhiêu giá trị nguyên của tham số m thuộc đoạn   1;20 để phương trình 2 13 2 4 2 x m x x x x      có nghiệm. A. 4. B. 18. C. 19. D. 20. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... DẠNG TOÁN 3: PHƯƠNG TRÌNH CHỨA ẨN TRONG CĂN BẬC HAI. 1. Phương pháp. Để giải phương trình chứa ẩn dưới dấu căn ta tìm cách để khử dấu căn, bằng cách:  Nâng luỹ thừa hai vế ( ) ( ) ( ) ( ) ( ) 0 ( ( ) 0) f x g x f x g x f x hay g x       ( ) ( ) f x g x     2 ( ) ( ) ( ) 0 f x g x gx         Phân tích thành tích.  Đặt ẩn phụ. 2. Bài tập minh họa. Loại 1: Bình phương hai vế của phương trình.  Bài tập 9. Giải các phương trình sau a). 2 2 4 2 x x x     b). 2 5 4 xx    c). 2 2 3 4 0 xx     d). 4 1 1 2 x x x      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 81 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 10. Giải các phương trình sau a). 2 3 1 1 xx    b). 2 2 1 3 1 0 x x x      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 82 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 11. Tìm m để phương trình 2 2 2 1 x mx x     có hai nghiệm phân biệt. Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Loại 2: Phân tích thành tích bằng cách nhân liên hợp. 1. Phương pháp Để trục căn thức ta nhân với các đại lượng liên hợp;     A B A B AB AB A B A B                       22 3 3 3 3 3 3 33 2 2 2 2 3 3 3 3 3 3 3 3 A B A A B B AB AB A A B B A A B B            Với A, B không đồng thời bằng không. 2. Bài tập minh họa. Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 83 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880  Bài tập 12. Giải các phương trình sau a). 2 1 2 3 x x x      b).   22 3 2 1 3 x x x x      c).     2 2 21 20 3 7 2 x x x    d). 3 3 2 2 xx    e). 22 3 3 8 15 2 x x x      f). 22 33 ( 2 31 3 ) x x xx      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 84 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Loại 3: Đặt ẩn phụ 1. Phương pháp. Phương trình có dạng Đặt ẩn phụ  2 , , ,... ax b x x  ,0 t ax b t     22 , ,... ax bx c ax bx    2 ,0 t ax bx c t      3 , ,... ax b ax b  3 t ax b           . f x g x f x g x       ,     f x g x C      t f x g x           2 , AA f x f x fx fx      A t f x fx       , mn f x f x   s t f x  với   ; s BCNN m n  2. Bài tập minh họa.  Bài tập 13. Giải các phương trình sau a). 22 11 31 xx    b). 2 ( 5)(2 ) 3 3 x x x x     c). 2 2 1 3 1 xx x xx    Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 85 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Giải các phương trình sau a). 2 4 1 4 6 1 0 x x x      b). 22 3 2 9 3 2 2 7 x x x x       c) 11 3 8 9 xx x x     d). 2 2 8 1 5 21 xx x x    Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 86 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Nhận xét: Phương trình có chứa     1 af x bf x  và     22 22 1 a f x b f x  thì ta đặt ẩn phụ là     1 t af x bf x  Bài tập 15. Giải các phương trình sau a). 22 1 1 2 x x x x       . b). 22 3 21 18 2 7 7 2 x x x x       . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 87 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài tập 16. Giải các phương trình sau a). 2 2 6 1 4 5 x x x     . b). 5 1 6 xx     . 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......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 17. Tìm m để phương trình sau có nghiệm a).   2 2 2 1 1 x m x x      (1) b). 2 4 3 1 1 2 1 x m x x      (2) Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 88 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Lưu ý: Khi giải bài toán bằng cách đặt ẩn phụ, đối với loại toán không chứa tham số thì có thể không nêu điều kiện(hoặc điều kiện "lỏng") của ẩn phụ vì sau khi tìm được nghiệm ẩn phụ rồi chúng ta phải thay lại để giải. Nhưng với bài toán chứa tham số thì chúng ta cần phải nêu điều kiện "chặt" đối với ẩn phụ. 3. Câu hỏi trắc nghiệm. Câu 26. Tập nghiệm S của phương trình 2 3 3 xx  là: A.   6;2 . S  B.   2. S  C.   6. S  D. . S  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 27. Tập nghiệm S của phương trình 2 42 xx    là: A.   0;2 . S  B.   2. S  C.   0. S  D. . S  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 89 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 28. Tổng các nghiệm của phương trình   2 2 2 7 4 x x x     bằng: A.0 . B.1. C. 2 . D.3 . Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu 29. Phương trình 2 42 2 2 xx x x    có tất cả bao nhiêu nghiệm? A.1. B. 2 . C. 3 . D. 5 . Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 30. Phương trình 4 22 23 x x     có tất cả bao nhiêu nghiệm? A.0 . B.1. C. 2 . D. 3 . Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 90 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 31. Có bao nhiêu giá trị nguyên của tham số m để phương trình 2 22 2 0 11 xx m xx        có đúng bốn nghiệm? A.0 . B.1. C. 2 . D. Vô số. Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu 32. Tìm tất cả các giá trị thực của tham số m để phương trình 2 2 11 2 1 0 x m x xx                  có nghiệm. A. 33 ;. 44 m     B. 3 ;. 4 m       C. 3 ;. 4 m        D. 33 ; ; . 44 m                Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 91 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 33. Tìm tất cả các giá trị thực của tham số m để phương trình 2 2 42 4 1 0 x x m xx          có đúng hai nghiệm lớn hơn 1. A. 8. m  B. 8 1. m    C. 0 1. m  D. 8. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 34. Tìm tất cả các giá trị thực của tham số m để phương trình     2 22 2 4 – 2 2 4 4 – 1 0 x x m x x m       có đúng hai nghiệm. A.   3;4 . m  B.     ;2 3 2 3; . m         C.     4; 2 3 . m      D. . m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 92 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 35. Tìm tất cả các giá trị thực của tham số m để phương trình 22 2 2 3 2 0 x mx m x m m m        có nghiệm. A.     ; 3 1; . m       B.   3 ; 3 ; . 2 m          C.   1; . m    D. 3 ;. 2 m       Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Loại 4: Đặt ẩn phụ không hoàn toàn  Bài tập 18. Giải phương trình a). 2 3 3 3 4 1 x x x     b).   22 3 1 3 1 x x x x      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Vậy phương trình ban đầu có hai nghiệm 1 x  và 2 x  Nhận xét: Trong lời giải trên ta thấy khó nhất là biến đổi phương trình ban đầu thành Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 93 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880   2 27 3 3 3 3 31 80 0 x x x x         để sau khi đặt ẩn phụ 3 tx  thì phương trình ẩn t có   2 18 93 x   (là bình phương của một nhị thức). Nếu ta tách không hợp lý thì không là bình phương của một nhị thức hoặc là một hằng số, trong trường hợp đó việc giải phương trình theo hướng trên là không thể thực hiện được. Vậy làm thế nào để tách được phương trình mà thỏa mãn các điều kiện trên và việc tách ra như thế có là duy nhất ? Để trả lời được câu hỏi này ta thực hiện theo các bước như sau: Bước 1: Viết       2 (1) 3 3 3 3 4 1 3 0 0 m x x x m x m m            Bước 2: Đặt   3 0 t x t    pt trở thành   22 3 3 4 1 3 0 mt t x m x m        Có     22 t 12 4 4 12 4 9 mx m m x m m f x          Bước 3: Tìm m sao cho     /2 / 12 0 12 0 27 4 27 1 0 0 f f m m m m m m m                      Đến đây việc giải pt như đã trình bày ở trên .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 19. Giải phương trình 22 60 24 5 5 10 x x x x      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 20. Giải phương trình       3 4 12 28 x x x x      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 94 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 21. Giải các phương trình sau a). 22 60 24 5 5 10 x x x x      . b).       3 4 12 28 x x x x      . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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.......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 22. Giải các phương trình sau a). 2 2 2 3 6 7 5 10 14 4 2 x x x x x x         . b). 2 5 2 3 2 5 2 2 2 x x x x         Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 95 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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DẠNG 4. Đưa về hệ phương trình  Bài tập 23. Giải phương trình a).   2 2 1 2 2 ( 1) 0 x x x     b). 32 10 1 3( 2) xx    c). 2 4 1 3 1 2 1 x x x       Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 96 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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.......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 24. Giải các phương trình sau a). 22 4 5 1 2 1 9 3 x x x x x        . b). 3 2 3 2 1 2 3 x x x x       . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 97 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 5. Bài tập luyện tập. Bài 10. Giải các phương trình sau a). 2 1 3 1 xx    b). 3 44 x x x    c). 4 4 2 3 1 1 x x x x      d). 2 2 6 1 1 x x x     e). 2 2 3 9 4 x x x     f). 2 77 xx    Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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.......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 11. Giải các phương trình sau: a). 22 12 5 3 5 x x x      b). 3 2 2 2 3 8 2 15 x x x      c). 2 3 5 1 9 2 3 1 x x x x       d). 2 3 6 7 1 x x x      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 98 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 12. Giải các phương trình sau a). 22 2 x x x x     b).   2 2 2 1 1 x x x     c). 13 2(3 2) 3 42 0 x x x      d). 22 2 22 2 24 0 x x x x        e). 11 2 13 x x xx      f). 2 4 1 4 6 1 0 x x x      g). 2 1 2 3 1 x x x x x     h). 3 2 4 2 21 x x x x     Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 99 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 13. Giải các phương trình sau a). 2 4 22 3 2 21 x x x     b).     2 1 5 3 3 4 x x x     c). 2 51 2 3 58 110 x x x     d). 2 3 1 2 6 x x x x     Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 14. Tìm m để phương trình     3 2 2 2 2 5 6 7 3 3 0 x m x m m x m         (*) có ba nghiệm dương phân biệt. Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 100 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 15. Cho phương trình   42 1 4 1 0 m x x     (*). Tìm m để a). Phương trình (*) có nghiệm b). Phương trình (*) có bốn nghiệm phân biệt Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 16. Cho phương trình 4 3 2 4 3 14 0 x x x x m      a). Giải phương trình khi 6 m  b). Tìm m để phương trình có nghiệm. Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 101 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Chú ý: Phương trình trên là phương trình có thể đưa về dạng     2 22 0 A x ax B x ax C      và cách giải là đặt 2 t x ax  và đưa về phương trình bậc hai 2 0 At Bt C    . Bài 17. Tìm m để phương trình : 2 ( 1)( 3)( 5) x x x m     có nghiệm. Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 18. Tìm m để phương trình : 43 48 x x x m    có bốn nghiệm phân biệt. 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 102 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... A–LÝ THUYẾ T I. Phương trình bậc nhất hai ẩn. 1. Định nghĩa: phương trình bậc nhất hai ẩn , xy có dạng tổng quát là   1 ax by c  trong đó ,, abc là các hệ số, với điều kiện a và b không đồng thời bằng 0. 2. Nhận xét: Khi 0 ab  ta có phương trình 0 0 . x y c   Nếu 0 c  thì phương trình này vô nghiệm.  Nếu 0 c  thì mọi cặp số   00 ; xy đều là nghiệm. Khi 0, b  phương trình ax by c  trở thành   2 ac yx bb     Cặp số   00 ; xy là một nghiệm của phương trình   1 khi và chỉ khi điểm   00 ; M x y thuộc đường thẳng   2. Tổng quát, người ta chứng minh được rằng phương trình bậc nhất hai ẩn luôn luôn có vô số nghiệm. Biểu diễn hình học tập nghiệm của phương trình của phương trình   1 là một đường thẳng trong mặt phẳng tọa độ . Oxy II. Hệ hai phương trình bậc nhất hai ẩn. 1. Định nghĩa: Hệ phương trình bậc nhất hai ẩn là hệ phương trình có dạng 1 1 1 2 2 2 2 1 1 2 2 2 2 2 ( 0, 0) ax by c a b a b a x b y c          Trong đó , xy là hai ẩn; các chữ số còn lại là hệ số. Nếu cặp số   00 ; xy đồng thời là nghiệm của cả hai phương trình của hệ thì   00 ; xy được gọi là một nghiệm của hệ phương trình   3. Giải hệ phương trình   3 là tìm tập nghiệm của nó. 2. Giải và biện luận hệ hai phương trình bậc nhất hai ẩn: Để giải hệ phương trình bậc nhất hai ẩn ta có thể dùng các cách giải đã biết ở chương trình lớp 9 như: Phương pháp thế. Phương pháp cộng đại số. Phương pháp đặt ẩn phụ. Trong chương trình lớp 10, để biện luận số nghiệm của hệ phương trình bậc nhất hai ẩn ta sử dụng định thức Cramer. Bước 1. Tính các định thức: 11 1 2 2 1 22 .. ab D a b a b ab    11 1 2 2 1 22 . x cb D cb c b cb    11 1 2 2 1 22 . y ac D ac a c ac    Bước 2. Xét định thức Kết quả 0 D  Hệ có nghiệm duy nhất ; y x D D xy DD     0 D  0 x D  hoặc 0 y D  Hệ vô nghiệm §B À I 4. H Ệ PH ƯƠNG TRÌNH B ẬC NH ẤT NHI ỀU ẨN Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 103 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 3. Ví dụ minh họa.  Ví dụ 1. Giải các hệ phương trình sau: a) 5 4 3 . 7 9 8 xy xy      b). 2 4 1 . 2 4 2 5 xy xy          Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Ví dụ 2. Giải các hệ phương trình sau: a). 65 3 9 10 1 xy xy          b). 62 3 22 34 1 22 x y x y x y x y              Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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.......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 0 xy DD  Hệ có vô số nghiệm Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 104 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 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III. Hệ phương trình bậc nhất ba ẩn 1. Định nghĩa: Hệ phương trình bậc nhất ba ẩn là hệ phương trình có dạng tổng quát 1 1 1 1 2 2 2 2 3 3 3 3 a x b y c z d a x b y c z d a x b y c z d               Trong đó ,, x y z là ba ẩn; các chữ còn lại là các hệ số. Mỗi bộ ba số   0 0 0 ;; x y z nghiệm đúng của ba phương trình của hệ được gọi là một nghiệm của hệ phương trình   4. Nếu cặp số   00 ; xy đồng thời là nghiệm của cả hai phương trình của hệ thì   00 ; xy được gọi là một nghiệm của hệ phương trình   3. Giải hệ phương trình   3 là tìm tập nghiệm của nó. 2. Giải và biện luận hệ hai phương trình bậc nhất hai ẩn: Để giải hệ phương trình bậc nhất ba ẩn, nguyên tắc chung là khử bớt ẩn để đưa về các phương trình hay hệ phương trình có số ẩn ít hơn. Để khử bớt ẩn, ta cũng có thể dùng các phương pháp cộng đại số, phương pháp thế như đối với hệ phương trình bậc nhất hai ẩn 3. Ví dụ minh họa.  Ví dụ 3. Giải hệ phương trình 3 3 1 (1) 2 2 (2) 2 2 3 (3) x y z x y z x y z                Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 105 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880  Ví dụ 4. Giải hệ phương trình 2 3 2 35 51 x y z x y z xy             Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... IV. Hệ phương trình gồm một phương trình bậc nhất và bậc 2. 1. Phương pháp. Sử dụng phương pháp thế  Từ phương trình bậc nhất rút một ẩn theo ẩn kia.  Thế vào phương trình bậc hai để đưa về phương trình bậc hai một ẩn.  Số nghiệm của hệ tuỳ theo số nghiệm của phương trình bậc hai này. 2. Ví dụ minh họa.  Ví dụ 5. Giải các hệ phương trình sau a). 2 4 2 5 0 y x x xy        b). 3 4 1 0 3( ) 9 xy xy x y          Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Giải các hệ phương trình sau Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 106 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 a). 3 2 23 x xy xy      b). 2 2 ( 1) 3 0 5 ( ) 1 0 x x y xy x              Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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V. Hệ phương trình đối xứng. 1. Hệ đối xứng loại 1 Định nghĩa: hệ phương trình đối xứng loại 1 là hệ phương trình có dạng:   ( , ) 0 ( , ) 0 f I xy g x y      với     ;; f x y f y x  và     ;; g x y g y x  . (Có nghĩa là khi ta hoán vị giữa x và y thì   ; f x y và   ; g x y không thay đổi). Cách giải  Đặt S x y  và . P x y  .  Đưa hệ phương trình   I về hệ   I  với các ẩn là S và P .  Giải hệ   I  ta tìm được S và P .  Tìm nghiệm   ; xy bằng cách giải phương trình: 2 0 X SX P    .  Ví dụ 7. Giải các hệ phương trình sau a). 22 2 3 2 6 x xy y xy           b). 22 33 30 35 x y xy xy        . 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 107 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 2. Hệ đối xứng loại 2 Định nghĩa: Hệ phương trình đối xứng loại 2 là hệ phương trình có dạng:   II ( , ) 0 (1) ( , ) 0 (2) f x y f y x      (Có nghĩa là khi hoán vị giữa x và y thì   1 biến thành   2 và ngược lại). Cách giải  Trừ   1 và   2 vế theo vế ta được:   II  ( , ) ( , ) 0 (3) ( , ) 0 f x y f y x f x y       Biến đổi   3 về phương trình tích:   3  ( ). ( , ) 0 x y g x y   ( , ) 0 xy g x y      .  Như vậy   II  ( , ) 0 ( , ) 0 ( , ) 0 f x y xy f x y g x y                 .  Giải các hệ phương trình trên ta tìm được nghiệm của hệ   II . Nhận xét: Hệ phương trình đối xứng loại 1, 2 nếu có nghiệm là   00 ; xy thì   00 ; yx cũng là một nghiệm của nó.  Ví dụ 8. Giải các hệ phương trình sau a). 3 3 2 2 x x y y y x        b). 2 3 2 2 3 2 32 32 y x x x x y y y            c). 2 2 2 2 2 3 2 3 x x y y y x            Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 108 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Tìm m để hệ phương trình 22 6 22 x xy y m x xy y m           có nghiệm duy nhất. 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 109 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Ví dụ 10. Cho   ; xy là nghiệm của hệ phương trình 2 2 2 21 2 2 3 x y m x y m m           . Tìm m để xy nhỏ nhất. 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B. CÁC DẠNG TOÁN VÀ PHƯƠNG PHÁP GIẢI. DẠNG TOÁN 1: GIẢI HỆ PHƯƠNG TRÌNH BẬC NHẤT HAI ẨN, BA ẨN. 1. Phương pháp Để giải hệ phương trình bậc nhất hai ẩn ta có thể dùng các cách như: Phương pháp thế. Phương pháp cộng đại số. Phương pháp đặt ẩn phụ. Phương pháp sử dụng định thức Crammer. Nhận xét: đối với bài toán biện luận tham số m thì dùng định thức Cramer rất nhanh. 2. Bài tập minh họa.  Bài tập 1. Giải các hệ phương trình sau a). 2 11 5 4 8 xy xy      c). 31 5 2 3 xy xy        L ời gi ải .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 110 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 2. Giải các hệ phương trình sau: a). ( 3)( 5) ( 2)( 5) x y xy x y xy          b). 2 21 xy xy          c). 3( ) 7 55 3 xy xy xy yx              Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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.......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 3. Giải các hệ phương trình sau: a). 6 3 2 5 11 4 2 4 2 11 xy yx xy yx              b). 2 6 3 1 5 5 6 4 1 1 xy xy              Lời giải .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 111 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 4. Giải các hệ phương trình sau: a) 31 2 2 5 2 3 0 x y z x y z x y z               c) 328 26 36 x y z x y z x y z               d) 3 2 7 2 4 3 8 35 x y z x y z x y z                 Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 112 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 3. Bài tập rèn luyện.  Bài 1. Giải các hệ phương trình sau a). 57 3 2 4 xy xy      b). 1 1 5 8 1 1 3 8 x y x y x y x y            c). 29 3 2 17 x y x y x y x y              d). 4 3 8 3 5 6 x y x y x y x y              Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 113 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 4. Câu hỏi trắc nghiệm. Câu 1. Cặp số   ; xy nào sau đây không là nghiệm của phương trình 2 3 5  xy ? A.   5 ; ; 0 2     xy . B.     ; 1; 1  xy . C.   5 ; 0; 3     xy . D.     ; 2; 3    xy . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 2. Phương trình 3 2 5 0 xy    nhận cặp số nào sau đây là nghiệm A.   2; 3  . B.   1; 1  . C.   3;2 . D.   1;1 . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 3. Tìm nghiệm của hệ phương trình 2 3 0 42 xy xy          . A.     ; 2;1 xy  . B.   10 1 ;; 77 xy     . C.   10 1 ;; 77 xy     . D.     ; 2; 1 xy    . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 4. Bộ     2; 1 ; 1; ; x y z   là nghiệm của hệ phương trình nào sau đây? A. 3 2 3 26 5 2 3 9 x y z x y z x y z                . B. 21 2 6 4 6 25 x y z x y z xy              . C. 31 2 0 x y z x y z x y z               . D. 2 26 10 4 2 x y z x y z x y z                . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 114 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Câu 5. Hệ phương trình 23 13 32 12 xy xy          có nghiệm là A. 1 2 x  ; 1 3 y  . B. 1 2 x  ; 1 3 y  . C. 1 2 x  ; 1 3 y  . D. 1 2 x  ; 1 3 y  . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 6. Hệ phương trình nào dưới đây vô nghiệm? A. 25 2 3 1 xy xy      . B. 31 13 1 22 xy xy          . C. 31 11 33 xy xy           . D. 32 5 xy xy      . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 7. Nghiệm của hệ phương trình 11 25 3 2 24 x y z x y z x y z               là: A.     5;3;3 . ;; x y z  B.     4;5;2 . ;; x y z  C.     2;4;5 . ;; x y z  D.     3;5;3 . ;; x y z  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 115 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Câu 8. Nghiệm của hệ phương trình 21 22 23 xy yz zx         là: A. 0 1. 1 x y z         B. 1 1. 0 x y z         C. 1 1. 1 x y z         D. 1 0. 1 x y z         Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 9. Bộ     2; 1;1 ;; x y z   là nghiệm của hệ phương trình nào sau đây ? A. 3 2 3 2 6 . 5 2 3 9 x y z x y z x y z                B. 21 2 6 4 6. 25 x y z x y z xy              C. 31 2. 0 x y z x y z x y z               D. 2 2 6 . 10 4 2 x y z x y z x y z                Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 10. Bộ     1 ; ; ; 1;0 x y z  là nghiệm của hệ phương trình nào sau đây ? A. 2 3 6 10 0 5. 4 17 x y z x y z yz                 B. 72 5 1 . 20 x y z x y z x y z                 C. 21 2. 2 x y z x y z x y z                 D. 22 4. 45 x y z x y z x y z                 Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 11. Gọi   00 ;; o x y z là nghiệm của hệ phương trình 3 3 1 22 2 2 3 x y z x y z x y z                . Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 116 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Tính giá trị của biểu thức 2 2 2 0 0 0 . P x y z    A. 1. P  B. 2. P  C. 3. P  D. 14. P  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 12. Gọi   00 ;; o x y z là nghiệm của hệ phương trình 11 25 3 2 24 x y z x y z x y z               . Tính giá trị của biểu thức 0 0 0 . P x y z  A. 40. P  B. 40. P  C. 1200. P  D. 1200. P  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 13. Hệ phương trình 22 22 7 3 x y xy x y xy            có tất cả các nghiệm là A.     ; 1; 2 ; xy        ; 2; 1 ; xy        ; 1;2 ; xy      ; 2; 1 xy . B.     ; 1; 2 ; xy        ; 2; 1 xy    . C.     ; 1;2 ; xy      ; 2;1 xy  . Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 117 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 D.     ; 1; 2 ; xy        ; 2; 1 xy    ;     ; 1;2 ; xy      ; 2;1 xy  . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 14. Nghiệm của hệ phương trình 41 5 2 52 3 2 xy xy            là A.     ; 3;11 xy  . B.     ; 3;1 xy  . C.     ; 13;1 xy  . D.     ; 3;1 xy  . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 15. Hệ phương trình 2 2 3 3 x x y y y x        có bao nhiêu nghiệm? A. 3 . B. 2 . C. 1. D. 4 . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 16. Hệ phương trình       2 2 2 2 2 2 5 4 6 4 4 0 1 23 2                 x y x y x xy y xy xy có một nghiệm   00 ; xy . Khi đó 2 00  P x y có giá trị là Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 118 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 A. 1. B. 17 16 . C. 3 . D. 2 . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 17.Cho hệ phương trình 22 2 2 8 3 12 9 4 18 6 7 2 3 1 0 x xy x y y x y x x y                  có nghiệm là   ; ab . Khi đó giá trị biểu thức 22 54 T a b  A. 24 T  . B. 21 T  . C. 5 T  . D. 4 T  . 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 119 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Câu 18. Các nghiệm của hệ 22 3 2 16 2 4 33 xy x y x y x y           là A.     ; 3 3; 2 3 ; xy          ; 3 3; 2 3 xy      . B.     ; 3; 2 ; xy        ; 3;2 xy  C.     ; 3 3; 3 3 ; xy          ; 2 3; 2 3 xy      . D.     ; 3;3 ; xy      ; 2;2 xy  . 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Câu 19. Cho   ; xy với x , y nguyên là nghiệm của hệ phương trình     2 2 71 12 2 xy y x y x x y          thì tích xy bằng A. 1. B. 2 . C. 3 . D. 4 . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 120 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... DẠNG TOÁN 2: GIẢI VÀ BIỆN LUẬN HỆ HAI PHƯƠNG TRÌNH BẬC NHẤT HAI ẨN. 1. Phương pháp. Sử dụng định thức: Tính ,, xy D D D Nếu 0 D  thì hệ có nghiệm duy nhất  ;; y x D D xy DD     Nếu 0 D  thì ta xét , xy DD  Với 0 0 x y D D      khi đó phương trình vô nghiệm.  Với 0 xy DD  thì hệ phương trình có vô số nghiệm tập nghiệm của hệ phương trình là tập nghiệm của một trong hai phương trình có trong hệ. 2. Bài tập minh họa.  Bài tập 5. Giải và biện luận hệ phương trình: 2 46 mx y m x my m        Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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.......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 6. Giải và biện luận hệ phương trình sau:     11 11 a x by b x ay            Lời giải Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 121 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... 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.........................................................................................................................................................................................................  Bài tập 7. Tìm m để hệ vô nghiệm     2 2 3 1 3 2 m x m y m x y y            Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 122 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 8. Tìm giá trị của b sao cho a  thì hệ phương trình   2 2 1 x ay b ax a y b          có nghiệm. Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .........................................................................................................................................................................................................  Bài tập 9. Tùy vào m hãy tìm giá trị lớn nhất nhỏ nhất của biểu thức sau:       22 ; 2 1 2 P x y x my mx y      Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 123 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 3. Bài tập luyện tập. Bài 1. Giải và biện luận các hệ phương trình sau: a). 22 3 mx y m xy      b). 22 ( 1) 2 1 2 m x y m m x y m m           Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Bài 2. Tìm m để hệ phương trình ( 1) 8 4 ( 3) 3 1 m x y m mx m y m           có nghiệm duy nhất Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 124 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 3. Tìm m để hệ phương trình   41 6 2 3 x my m m x y m              có vô số nghiệm: Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 4. Tùy theo giá trị của m , hãy tìm giá trị nhỏ nhất của biểu thức :       22 ; 2 2 3 P x y mx y m x y       Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Bài 5. Cho hệ phương trình: (2 1) 3 3 2 ( 3) ( 1) 2 m x y m m x m y m            . a). Tìm m để hệ có nghiệm. b). Tìm m để hệ có nghiệm duy nhất   ; xy thỏa mãn 2 xy  . c). Tìm m để hệ có nghiệm duy nhất   ; xy sao cho 22 3 P x y  nhỏ nhất . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 125 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Bài 6. Giải và biện luận hệ: 11 12 m x y m x m y             Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... 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Câu hỏi trắc nghiệm. Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 126 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Câu 20. Cho hệ phương trình 1 1      x my mx y   I , m là tham số. Mệnh đề nào sai? A. Hệ   I có nghiệm duy nhất 1    m . B. Khi 1  m thì hệ   I có vô số nghiệm. C. Khi 1  m thì hệ   I vô nghiệm. D. Hệ   I có vô số nghiệm. Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 21. Tìm điều kiện của tham số m để hệ phương trình 1 mx y m x my         có nghiệm duy nhất. A. 1 m  . B. 1 m  . C. 1 m  . D. 1 m  . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 22. Cho hệ phương trình 21 3 4 1 x y m x y m          . Giá trị m thuộc khoảng nào sau đây để hệ phương trình có nghiệm duy nhất   00 ; xy thỏa mãn 00 2 3 1 xy  ? A.   5; 9 m  . B.   5;1 m . C.   0; 3 m  . D.   4;1 m . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 23. Có bao nhiêu giá trị m nguyên dương để hệ phương trình 3 29 mx y x my      có nghiệm duy nhất   ; xy sao cho biểu thức 3 A x y  nhận giá trị nguyên A. 4. B. 2. C. 3. D. 1. Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 127 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 24. Tìm giá trị thực của tham số m để hệ phương trình 2 3 4 0 3 1 0 2 5 0 xy xy mx y m               có duy nhất một nghiệm. A. 10 . 3 m  B. 10. m  C. 10. m  D. 10 . 3 m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 25. Tìm giá trị thực của tham số m để hệ phương trình 1 1 1 mx y my z x mz         vô nghiệm. A. 1. m  B. 0. m  C. 1. m  D. 1. m  Lời giải. .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 128 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 26. Khi hệ phương trình   21 2 2 2 41 x my z x my z x m y z                có nghiệm   ;; x y z với 0 4 3 m m        , giá trị 2017 2018 2017 T x y z    là A. 2017 T  . B. 2018 T  . C. 2017 T  . D. 2018 T   . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Câu 27. Cho hệ phương trình 2 2 2 2 42 xy x y xy m m        . Tìm tất cả các giá trị của m để hệ trên có nghiệm. A. 1 ;1 2     . B.   1;  . C.   0;2 . D. 1 ; 2        . 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Câu 28. Hệ phương trình 2 22 3 4          x xy y xy m có nghiệm khi A. 1 1      m m . B. 1  m . C. 1  m . D. 1  m . Lời giải .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... .......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... ......................................................................................................................................................................................................... Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 129 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Câu 29. Một đoàn xe tải chở 290 tấn xi măng cho một công trình xây đập thủy điện. Đoàn xe có 57 chiếc gồm ba loại, xe chở 3 tấn, xe chở 5 tấn và xe chở 7,5 tấn. Nếu dùng tất cả xe 7,5 tấn chở ba chuyến thì được số xi măng bằng tổng số xi măng do xe 5 tấn chở ba chuyến và xe 3 tấn chở hai chuyến. Hỏi số xe mỗi loại ? A. 18 xe chở 3 tấn, 19 xe chở 5 tấn và 20 xe chở 7,5 tấn. B. 20 xe chở 3 tấn, 19 xe chở 5 tấn và 18 xe chở 7,5 tấn. C. 19 xe chở 3 tấn, 20 xe chở 5 tấn và 18 xe chở 7,5 tấn. D. 20 xe chở 3 tấn, 18 xe chở 5 tấn và 19 xe chở 7,5 tấn. 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Câu 30. Có ba lớp học sinh 10 , 10 , 10 A B C gồm 128 em cùng tham gia lao động trồng cây. Mỗi em lớp 10A trồng được 3 cây bạch đàn và 4 cây bàng. Mỗi em lớp 10B trồng được 2 cây bạch đàn và 5 cây bàng. Mỗi em lớp 10C trồng được 6 cây bạch đàn. Cả ba lớp trồng được là 476 cây bạch đàn và 375 cây bàng. Hỏi mỗi lớp có bao nhiêu học sinh ? A. 10A có 40 em, lớp 10B có 43 em, lớp 10C có 45 em. B. 10A có 45 em, lớp 10B có 43 em, lớp 10C có 40 em. C. 10A có 45 em, lớp 10B có 40 em, lớp 10C có 43 em. D. 10A có 43 em, lớp 10B có 40 em, lớp 10C có 45 em. 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Câu 31. Hai bạn Vân và Lan đi mua trái cây. Vân mua 10 quả quýt, 7 quả cam với giá tiền là 17800. Lan mua 12 quả quýt, 6 quả cam hết 18000. Hỏi giá tiền mỗi quả quýt, quả cam là bao nhiêu? A. Quýt 1400, cam 800 . B. Quýt 700 , cam 200 . C. Quýt 800 , cam 1400. D. Quýt 600 , cam 800 . 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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III–Phương trình và Hệ Phương Trình 130 Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880 Câu 32. Một xe hơi khởi hành từ Krông Năng đi đến Nha Trang cách nhau 175km. Khi về xe tăng vận tốc trung bình hơn vận tốc trung bình lúc đi là 20 km/giờ. Biết rằng thời gian dùng để đi và về là 6 giờ; vận tốc trung bình lúc đi là A. 60 km/giờ. B. 45 km/giờ. C. 55 km/giờ. D. 50 km/giờ. 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