Đáp án + Giải thích các bước giải:
Bài 1
a ) P = x3 – x = x ( x2 – 1 )
Để P = 0 thì có hai trường hợp xảy ra
Th1 : x = 0
Th2 : x2 – 1 = 0 → x2 = 1 → x = ± 1
b) Q = x( x – 2 ) ( x + 2 ) – ( 2x + 3 ) ( x – 1 ) – 2x2
= x ( x2 – 4 ) – ( 2x2 – x – 3 ) – 2x2
= x3 – 4x – 2x2 + x + 3 – 2x2
= x3 – 4x2 – 3x + 3
x = ½ → Q = ( ½ )3 – 4( ½ )2 – 3. ½ + 3
= 1/8 – 1 – 3/2 + 3 = 1/8 – 12/8 + 2
= -11/8 + 2 = 5/8
c) M = Q – P = x3 – 4x2 – 3x + 3 – x ( x2 – 1 )
= x3 – 4x2 – 3x + 3 – x3 + x = – 4x2 + 2x + 3
= – ( 4x2 – 2x – 3 ) = – [ (2x)2 – 2x – ¼ – 11/4 ] = –[ ( 2x – ½ )2 – 11/4 ]
= – ( 2x – ½ )2 + 11/4 ≤ 11/4
Max M = 11/4 → 2x – ½ = 0 → x = ¼
Câu 2 :
a) x2 – xy + 3x – 3y = x( x – y) + 3( x – y ) = ( x + 3 ) ( x – y)
b) x3 – 4x2y + 4xy2 + 9x = x ( x2 – 9 ) – 4xy ( x – y) ( Câu này sai đề nhé bạn )
c) 2x2 – 7x + 3 = 2 ( x2 – 7/2x + 3/2 ) = 2 ( x2 – 7/2 x + 49/16 – 25/16 )
= 2 [ ( x – 7/4 )2 – 25/16 ]
