Chuyên đề: Các bài tập nâng cao trong rút gọn đa thức
A. Lý thuyết
- Nắm vững các dạng rút gọn, các hằng đẳng thức, quy đồng đặt nhân tử chung,...
B. Bài tập
I. Bài tập minh họa
Câu 1. Cho biểu thức \[P=\frac{(\frac{a}{b}+\frac{b}{a}+1){{(\frac{1}{a}-\frac{1}{b})}^{2}}}{\frac{{{a}^{2}}}{{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}}-(\frac{a}{b}+\frac{b}{a})}\] với a > 0, b > 0, a ≠ b.
- Chứng minh \[P=\frac{1}{ab}\]
- Giả sử a, b thay đổi sao cho \[4a+b+\sqrt{ab}=1\]. Tìm giá trị nhỏ nhất của P.
Giải
Ta có:
\[P=\frac{(\frac{a}{b}+\frac{b}{a}+1){{(\frac{1}{a}-\frac{1}{b})}^{2}}}{\frac{{{a}^{2}}}{{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}}-(\frac{a}{b}+\frac{b}{a})}=\frac{\left( \frac{{{a}^{2}}}{ab}+\frac{{{b}^{2}}}{ab}+\frac{ab}{ab} \right){{\left( \frac{a-b}{ab} \right)}^{2}}}{\frac{{{a}^{4}}}{{{a}^{2}}{{b}^{2}}}+\frac{{{b}^{4}}}{{{a}^{2}}{{b}^{2}}}-\left( \frac{{{a}^{3}}b}{{{a}^{2}}{{b}^{2}}}+\frac{a{{b}^{3}}}{{{a}^{2}}{{b}^{2}}} \right)}=\frac{\frac{\left( {{a}^{3}}-{{b}^{3}} \right)\left( a-b \right)}{{{a}^{3}}{{b}^{3}}}}{\frac{\left( {{a}^{3}}-{{b}^{3}} \right)\left( a-b \right)}{{{a}^{2}}{{b}^{2}}}}=\frac{1}{ab}\]
Câu 2.
Cho hai số thực a , b thỏa điều kiện ab = 1, a +b ¹ 0 . Tính giá trị của biểu thức:
\[P=\frac{1}{{{(a+b)}^{3}}}(\frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}})+\frac{3}{{{(a+b)}^{4}}}(\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}})+\frac{6}{{{(a+b)}^{5}}}(\frac{1}{a}+\frac{1}{b})\]
Giải
Với ab = 1 , a + b ¹ 0, ta có:
\[P=\frac{1}{{{(a+b)}^{3}}}(\frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}})+\frac{3}{{{(a+b)}^{4}}}(\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}})+\frac{6}{{{(a+b)}^{5}}}(\frac{1}{a}+\frac{1}{b})\]\[=\frac{{{a}^{3}}+{{b}^{3}}}{{{\left( a+b \right)}^{3}}}+\frac{3\left( {{a}^{2}}+{{b}^{2}} \right)}{{{\left( a+b \right)}^{4}}}+\frac{6\left( a+b \right)}{{{\left( a+b \right)}^{5}}}=\frac{{{a}^{2}}+{{b}^{2}}-1}{{{\left( a+b \right)}^{2}}}+\frac{3\left( {{a}^{2}}+{{b}^{2}} \right)}{{{\left( a+b \right)}^{4}}}+\frac{6\left( a+b \right)}{{{\left( a+b \right)}^{5}}}\]\[=\frac{\left( {{a}^{2}}+{{b}^{2}}-1 \right){{\left( a+b \right)}^{2}}+3\left( {{a}^{2}}+{{b}^{2}} \right)+6}{{{\left( a+b \right)}^{4}}}=\frac{\left( {{a}^{2}}+{{b}^{2}}-1 \right)\left( {{a}^{2}}+{{b}^{2}}+2 \right)+3\left( {{a}^{2}}+{{b}^{2}} \right)+6}{{{\left( a+b \right)}^{4}}}\]\[=\frac{{{\left( {{a}^{2}}+{{b}^{2}}+2 \right)}^{2}}}{{{\left( a+b \right)}^{4}}}=\frac{{{\left( {{a}^{2}}+{{b}^{2}}+2ab \right)}^{2}}}{{{\left( a+b \right)}^{4}}}=1\]
Vậy P = 1, với ab = 1 , a+b ¹ 0.
Câu 3.
Cho các số thực dương a, b.Chứng minh rằng
\[\frac{\frac{{{(a-b)}^{3}}}{{{(\sqrt{a}-\sqrt{b})}^{3}}}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}+\frac{3a+3\sqrt{ab}}{b-a}=0\]
Giải
\[Q=\frac{\frac{{{(a-b)}^{3}}}{{{(\sqrt{a}-\sqrt{b})}^{3}}}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}+\frac{3a+3\sqrt{ab}}{b-a}\]\[=\frac{\frac{{{(\sqrt{a}-\sqrt{b})}^{3}}.{{(\sqrt{a}+\sqrt{b})}^{3}}}{{{(\sqrt{a}-\sqrt{b})}^{3}}}-b\sqrt{b}+2a\sqrt{a}}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}-\frac{3\sqrt{a}(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}\]\[=\frac{a\sqrt{a}+3a\sqrt{b}+3b\sqrt{a}+b\sqrt{b}+2a\sqrt{a}}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}-\frac{3\sqrt{a}}{\sqrt{a}-\sqrt{b}}\]\[=\frac{3a\sqrt{a}+3a\sqrt{b}+3b\sqrt{a}-3a\sqrt{a}-3a\sqrt{b}-3b\sqrt{a}}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}\]\[=0(DPCM)\]
Câu 4.Cho các số dương x, y, z thỏa mãn các điều kiện x + y + z = 2 và x2 + y2 + z2 = 2.
Chứng minh rằng biểu thức sau không phụ thuộc vào x, y, z:
\[P=x\sqrt{\frac{\left( 1+{{y}^{2}} \right)\left( 1+{{z}^{2}} \right)}{1+{{x}^{2}}}}+y\sqrt{\frac{\left( 1+{{x}^{2}} \right)\left( 1+{{z}^{2}} \right)}{1+{{y}^{2}}}}+z\sqrt{\frac{\left( 1+{{y}^{2}} \right)\left( 1+{{x}^{2}} \right)}{1+{{z}^{2}}}}\]
Giải
Xét \[{{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\left( xy+x\text{z}+yz \right)\Rightarrow xy+yz+x\text{z}=\frac{{{\left( x+y+z \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}{2}\]
Thay x + y + z = 2 và x2 + y2 + z2 = 2 ta có xy + yz + zx = 1.
Thay 1 = xy + yz + zx ta có:
\[x\sqrt{\frac{\left( 1+{{y}^{2}} \right)\left( 1+{{z}^{2}} \right)}{1+{{x}^{2}}}}=x\sqrt{\frac{\left( xy+yz+z\text{x}+{{y}^{2}} \right)\left( xy+yz+z\text{x}+{{z}^{2}} \right)}{xy+yz+z\text{x}+{{x}^{2}}}}=x\sqrt{\frac{\left( y+z \right)\left( y+x \right)\left( z+y \right)\left( z+x \right)}{\left( x+y \right)\left( x+z \right)}}=x\left( y+z \right)\]
Tương tự ta có:
\[y\sqrt{\frac{\left( 1+{{z}^{2}} \right)\left( 1+{{x}^{2}} \right)}{1+{{y}^{2}}}}=y\left( z+x \right)\]
\[z\sqrt{\frac{\left( 1+{{y}^{2}} \right)\left( 1+{{x}^{2}} \right)}{1+{{z}^{2}}}}=z\left( y+x \right)\]
Cộng từng vế của ba đẳng thức trên ta có
\[P=xy+x\text{z}+yz+xy+z\text{x}+zy=2\left( xy+yz+z\text{x} \right)=2\]
Vậy biểu thức P không phụ thuộc vào x, y, z.
Câu 5.
Cho 3 số thực x, y, z thỏa mãn điều kiện: x + y + z = 0 và xyz ≠ 0.
Tính giá trị biểu thức \[P=\frac{{{x}^{2}}}{{{y}^{2}}+{{z}^{2}}-{{x}^{2}}}+\frac{{{y}^{2}}}{{{z}^{2}}+{{x}^{2}}-{{y}^{2}}}+\frac{{{z}^{2}}}{{{x}^{2}}+{{y}^{2}}-{{z}^{2}}}\]
Giải
Ta có
\[x+y+z=0\Leftrightarrow {{\left( y+z \right)}^{2}}={{\left( -x \right)}^{2}}\Leftrightarrow {{y}^{2}}+{{z}^{2}}-{{x}^{2}}=-2yz\]
Tương tự:
\[{{z}^{2}}+{{x}^{2}}-{{y}^{2}}=-2\text{zx}\]
\[{{x}^{2}}+{{y}^{2}}-{{z}^{2}}=-2yz\]
\[P=\frac{{{x}^{2}}}{-2yz}+\frac{{{y}^{2}}}{-2\text{zx}}+\frac{{{z}^{2}}}{-2yz}=\frac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}}{-2\text{x}yz}\]
Mà
\[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}={{\left( x+y \right)}^{3}}-3{{\text{x}}^{2}}y-3\text{x}{{y}^{2}}+{{z}^{3}}={{\left( -z \right)}^{3}}-3\text{x}y\left( x+y \right)+{{z}^{3}}=3\text{x}y\Rightarrow P=\frac{3\text{xyz}}{-2\text{x}yz}=-\frac{3}{2}\]
II. Bài tập tự luyện
Câu 1: Rút gọn các biểu thức:
a) A = $\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}$
b) B = $\left( \frac{1}{\text{x}-4}-\frac{1}{\text{x + 4}\sqrt{\text{x}}+4} \right).\frac{\text{x + 2}\sqrt{\text{x}}}{\sqrt{\text{x}}}$ ( với x > 0, x $\ne $ 4 ).
Câu 2: Rút gọn các biểu thức sau:
a) A = $\left( 2+\frac{3+\sqrt{3}}{\sqrt{3}+1} \right).\left( 2-\frac{3-\sqrt{3}}{\sqrt{3}-1} \right)$
b) B = $\left( \frac{\sqrt{\text{b}}}{\text{a - }\sqrt{\text{ab}}}\text{ - }\frac{\sqrt{\text{a}}}{\sqrt{\text{ab}}\text{ - b}} \right)\text{.}\left( \text{a}\sqrt{\text{b}}\text{ - b}\sqrt{\text{a}} \right)$ ( với a > 0, b > 0, a $\ne $b)
Câu 3: Cho biểu thức A = $\left( \frac{\sqrt{\text{a}}}{\sqrt{\text{a}}-1}-\frac{\sqrt{\text{a}}}{\text{a - }\sqrt{\text{a}}} \right):\frac{\sqrt{\text{a}}+1}{\text{a - 1}}$ với a > 0, a $\ne $ 1
a) Rút gọn biểu thức A.
b) Tìm các giá trị của a để A < 0.
Câu 4: a) Rút gọn biểu thức: A = $\left( \frac{3\sqrt{\text{x}}+6}{\text{x - 4}}+\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}-2} \right):\frac{\text{x - 9}}{\sqrt{\text{x}}-3}$ với $\text{x }\ge \text{ 0, x }\ne \text{ 4, x }\ne \text{ 9}$.
Câu 5: Rút gọn các biểu thức:
a) A = $3\sqrt{8}-\sqrt{50}-\sqrt{{{\left( \sqrt{2}-1 \right)}^{2}}}$
b) B = $\frac{2}{\text{x - 1}}.\sqrt{\frac{{{\text{x}}^{\text{2}}}\text{ - 2x + 1}}{4{{\text{x}}^{\text{2}}}}}$, với 0 < x < 1
Câu 6: 1) Rút gọn biểu thức:
$A\text{ }=\text{ }\left( \frac{1\text{ }-\text{ }a\sqrt{a}}{1\text{ }-\text{ }\sqrt{a}}\text{ }+\text{ }\sqrt{a} \right){{\left( \frac{1\text{ }-\text{ }\sqrt{a}}{1\text{ }-\text{ }a} \right)}^{2}}$ với a ≥ 0 và a ≠ 1.
Câu 7: Tính gọn biểu thức:
1) A = $\sqrt{20}\text{ - }\sqrt{45}\text{ + 3}\sqrt{18}\text{ + }\sqrt{72}$.
2) B = $\left( 1\text{ + }\frac{\text{a + }\sqrt{\text{a}}}{\sqrt{\text{a}}\text{ + 1}} \right)\left( \text{1 + }\frac{\text{a - }\sqrt{\text{a}}}{\text{ 1}\,\text{-}\,\,\sqrt{a}} \right)$ với a ≥ 0, a ≠ 1.
Câu 8: Cho biểu thức: P = $\left( \frac{\text{a}\sqrt{\text{a}}\text{ - 1}}{\text{a - }\sqrt{\text{a}}}\text{ - }\frac{\text{a}\sqrt{\text{a}}\text{ + 1}}{\text{a + }\sqrt{\text{a}}} \right)\text{ : }\frac{\text{a +}\text{2}}{\text{a - 2}}$ với a > 0, a ¹ 1, a ¹ 2.
1) Rút gọn P.
2) Tìm giá trị nguyên của a để P có giá trị nguyên.
Câu 9: Cho biểu thức
P = $\frac{\sqrt{\text{x}}\text{ + 1}}{\sqrt{\text{x}}\text{ - 2}}\text{ + }\frac{2\sqrt{\text{x}}}{\sqrt{\text{x}}\text{ + 2}}\text{ + }\frac{2\text{ + 5}\sqrt{\text{x}}}{\text{4 - x}}$ với x ≥ 0, x ≠ 4.
1) Rút gọn P.
2) Tìm x để P = 2.
Câu 10: Cho M = $\left( \frac{\sqrt{\text{x}}}{\sqrt{\text{x}}\text{ - 1}}\text{ - }\frac{1}{\text{x - }\sqrt{\text{x}}} \right)\text{ : }\left( \frac{1}{\sqrt{\text{x}}+1}\text{ + }\frac{2}{\text{x - 1}} \right)$ với $x>0,\,\,x\ne 1$.
a) Rút gọn M.
b) Tìm x sao cho M > 0.
Câu 11: Cho biểu thức: K = $\frac{\text{x}}{\sqrt{\text{x}}\text{ - }1}\text{ - }\frac{2\text{x - }\sqrt{\text{x}}}{\text{x - }\sqrt{\text{x}}}$ với x >0 và x$\ne $1
Rút gọn biểu thức K
Tìm giá trị của biểu thức K tại x = 4 + 2$\sqrt{3}$
Câu 12: Rút gọn các biểu thức:
1) $\sqrt{45}+\sqrt{20}-\sqrt{5}$.
2) $\frac{\text{x}+\sqrt{\text{x}}}{\sqrt{\text{x}}}+\frac{\text{x}-4}{\sqrt{\text{x}}+2}$ với x > 0.
Câu 13: Cho các biểu thức A = $\frac{5+7\sqrt{5}}{\sqrt{5}}+\frac{11+\sqrt{11}}{1+\sqrt{11}}\,\,\,,\,\,\,\,B=\sqrt{5}:\frac{5}{5+\sqrt{55}}$
a) Rút gọn biểu thức A.
b) Chứng minh: A - B = 7.
Câu 14: Rút gọn các biểu thức :
a) A = $\frac{2}{\sqrt{5}\text{ - 2}}\text{ - }\frac{2}{\sqrt{5}\text{ + 2}}$
b) B = $\left( \sqrt{\text{x}}\text{ - }\frac{1}{\sqrt{\text{x}}} \right)\text{ : }\left( \frac{\sqrt{\text{x}}\text{ - 1}}{\sqrt{\text{x}}}\text{ + }\frac{1\text{ - }\sqrt{\text{x}}}{\text{x + }\sqrt{\text{x}}} \right)$ với $x>0,\,\,x\ne 1.$
Câu 15: Cho biểu thức: P = \[\left( \frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}} \right)\left( \frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1} \right)\]
1) Rút gọn biểu thức P.
2) Tìm a để P > - 2
Câu 16. Rút gọn:
A = $(1-\sqrt{5})\cdot \frac{\sqrt{5}+5}{2\sqrt{5}}.$
B = $\left( 1+\frac{x+\sqrt{x}}{1+\sqrt{x}} \right)\left( 1+\frac{x-\sqrt{x}}{1-\sqrt{x}} \right)$ với $0\le x\ne 1$.
Câu 17. Cho biểu thức A = $\left( \frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}} \right):\left( \frac{1}{\sqrt{x}+1}+\frac{2}{x-1} \right)$
1) Rút gọn biểu thức A.
2) Tính giá trị của A khi $x=2\sqrt{2}+3$.
Câu 18: Cho biểu thức P = $\left( \frac{\text{1}}{\text{x + }\sqrt{\text{x}}}-\frac{1}{\sqrt{\text{x}}+1} \right):\frac{\sqrt{\text{x}}}{\text{x + 2}\sqrt{\text{x}}+1}$ với x > 0.Rút gọn biểu thức P.
Tìm các giá trị của x để P > $\frac{1}{2}$.
Câu 19: Rút gọn các biểu thức sau:
1) A = $\frac{1}{2}\sqrt{20}-\sqrt{80}+\frac{2}{3}\sqrt{45}$
2) B = $\left( 2+\frac{5-\sqrt{5}}{\sqrt{5}-1} \right).\left( 2-\frac{5+\sqrt{5}}{\sqrt{5}+1} \right)$
Câu 20: Cho biểu thức A = $\left( \frac{\sqrt{\text{a}}}{\sqrt{\text{a}}+1}-\frac{\sqrt{\text{a}}}{\text{a + }\sqrt{\text{a}}} \right):\frac{\sqrt{\text{a}}-1}{\text{a - 1}}$ với a > 0, a $\ne $ 1.
1) Rút gọn biểu thức A.
2) Tìm các giá trị của a để A < 0.
Câu 21: Cho biểu thức P = \[\left( \frac{1}{\sqrt{a}-3}+\frac{1}{\sqrt{a}+3} \right)\left( 1-\frac{3}{\sqrt{a}} \right)\] với a > 0 và a \[\ne \] 9.
a) Rút gọn biểu thức P
b) Tìm các giá trị của a để P > \[\frac{1}{2}\].
Câu 22. Rút gọn biểu thức P = $\frac{9\sqrt{a}-\sqrt{25a}+\sqrt{4{{a}^{3}}}}{{{a}^{2}}+2a}$ với $a>0$.
Câu 23: Tính:
a) \[A=\sqrt{20}-3\sqrt{18}-\sqrt{45}+\sqrt{72}\].
b) \[B=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\].
c) \[C=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\] với x > 1
Câu 24: 1) Rút gọn biểu thức: P = $(\sqrt{7}+\sqrt{3}-2)(\sqrt{7}-\sqrt{3}+2)$.
Câu 25: Cho biểu thức A =\[\left( 1-\frac{2\sqrt{a}}{a+1} \right):\left( \frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1} \right)\]. Rút gọn biểu thức A.
Câu 26: Rút gọn biểu thức: P = \[\sqrt{{{(\sqrt{a-1}+1)}^{2}}}+\sqrt{{{(\sqrt{a-1}-1)}^{2}}}\] với a > 1
Câu 27: Cho biểu thức: Q = \[{{\left( \frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}} \right)}^{2}}\left( \frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1} \right)\]. Tìm tất cả các giá trị của x để Q có nghĩa. Rút gọn Q.
Câu 28: Rút gọn A = \[\frac{\sqrt{{{x}^{2}}+6x+9}}{x+3}\] với $x\ne -3$.
Câu 29: a) Tính \[\sqrt{{{(1+\sqrt{5})}^{2}}}+\sqrt{{{(1-\sqrt{5})}^{2}}}\].
b) Giải phương trình: x2 + 2x - 24 = 0.
Câu 30: Cho biểu thức: P = \[\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}+1}{\sqrt{a}-3}+\frac{3+7\sqrt{a}}{9-a}\] với a > 0, a \[\ne \] 9.
a) Rút gọn.
b) Tìm a để P < 1.
Câu 31: Cho biểu thức: M = \[\frac{{{x}^{2}}-\sqrt{x}}{x+\sqrt{x}+1}-\frac{{{x}^{2}}+\sqrt{x}}{x-\sqrt{x}+1}+x+1\]
Rút gọn biểu thức M với $x\ge 0.$
Câu 32: Cho biểu thức: P = \[\frac{{{x}^{2}}+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\] với x > 0.Rút gọi biểu thức P.
Câu 33:
Tính: $\sqrt{48} - 2\sqrt{75} + \sqrt{108}$
Rút gọn biểu thức: P= $\left( \frac{1}{1 - \sqrt{x}} - \frac{1}{1 + \sqrt{x}} \right) . \left( 1 - \frac{1}{\sqrt{x}} \right)$ với x$\ne $1 và x >0
ĐÁP ÁN
Câu 1:$\text{ a) A = }\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}=\frac{\sqrt{3}\left( 1-\sqrt{2} \right)}{1-\sqrt{2}}-\frac{2\left( 1+\sqrt{2} \right)}{1+\sqrt{2}}=\sqrt{3}-2$
$\text{b) B = }\left( \frac{1}{\text{x}-4}-\frac{1}{\text{x + 4}\sqrt{\text{x}}+4} \right).\frac{\text{x + 2}\sqrt{\text{x}}}{\sqrt{\text{x}}}$ $\text{= }\left( \frac{1}{\left( \sqrt{\text{x}}-2 \right)\left( \sqrt{\text{x}}+2 \right)}-\frac{1}{{{(\sqrt{\text{x}}+2)}^{2}}} \right).\frac{\sqrt{\text{x}}\text{(}\sqrt{\text{x}}\text{ + 2)}}{\sqrt{\text{x}}}$
$\text{=}\frac{1}{\sqrt{\text{x}}-2}-\frac{1}{\sqrt{\text{x}}+2}=\frac{\left( \sqrt{\text{x}}+2 \right)-\left( \sqrt{\text{x}}-2 \right)}{\text{x - 4}}=\frac{4}{\text{x - 4}}$
Câu 2:
$\text{ a) A = }\left( 2+\frac{3+\sqrt{3}}{\sqrt{3}+1} \right).\left( 2-\frac{3-\sqrt{3}}{\sqrt{3}-1} \right)=\left( 2+\frac{\sqrt{3}\left( \sqrt{3}+1 \right)}{\sqrt{3}+1} \right)\left( 2-\frac{\sqrt{3}\left( \sqrt{3}-1 \right)}{\sqrt{3}-1} \right)=\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)=1.$
$\text{b) }\left( \frac{\sqrt{\text{b}}}{\text{a - }\sqrt{\text{ab}}}\text{ - }\frac{\sqrt{\text{a}}}{\sqrt{\text{ab}}\text{ - b}} \right)\text{.}\left( \text{a}\sqrt{\text{b}}\text{ - b}\sqrt{\text{a}} \right)=\left( \frac{\sqrt{\text{b}}}{\sqrt{\text{a}}\left( \sqrt{\text{a}}-\sqrt{\text{b}} \right)}\text{ - }\frac{\sqrt{\text{a}}}{\sqrt{\text{b}}\left( \sqrt{\text{a}}-\sqrt{\text{b}} \right)} \right)\text{.}\sqrt{\text{ab}}\left( \sqrt{\text{a}}\text{ - }\sqrt{\text{b}} \right)$
$=\frac{\sqrt{\text{b}}.\sqrt{\text{ab}}}{\sqrt{\text{a}}}-\frac{\sqrt{\text{a}}.\sqrt{\text{ab}}}{\sqrt{\text{b}}}=\text{b - a}\text{.}$
Câu 3:
$\mathbf{ }\text{ a) A =}\left( \frac{\sqrt{\text{a}}}{\sqrt{\text{a}}-1}-\frac{\sqrt{\text{a}}}{\sqrt{\text{a}}\text{(}\sqrt{\text{a}}\text{ - 1)}} \right):\frac{\sqrt{\text{a}}+1}{(\sqrt{\text{a}}\text{ - 1)(}\sqrt{\text{a}}+1)}=\left( \frac{\sqrt{\text{a}}}{\sqrt{\text{a}}-1}-\frac{\text{1}}{\text{(}\sqrt{\text{a}}\text{ - 1)}} \right).\left( \sqrt{\text{a}}-1 \right)=\sqrt{\text{a}}-1$
b) A < 0 .png)
Câu 4: a) A =$\left( \frac{3\sqrt{\text{x}}+6}{\text{x - 4}}+\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}-2} \right):\frac{\text{x - 9}}{\sqrt{\text{x}}-3}$
$=\left( \frac{3(\sqrt{\text{x}}+2)}{\left( \sqrt{\text{x}}-2 \right)\left( \sqrt{\text{x}}+2 \right)}+\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}-2} \right):\frac{\left( \sqrt{\text{x}}-3 \right)\left( \sqrt{\text{x}}+3 \right)}{\sqrt{\text{x}}-3}$
$=\left( \frac{3+\sqrt{\text{x}}}{\sqrt{\text{x}}-2} \right).\frac{1}{\sqrt{\text{x}}+3}=\frac{1}{\sqrt{\text{x}}-2}$, với $\text{x }\ge \text{ 0,}\,\,\text{x }\ne \text{ 4,}\,\,\text{x }\ne \text{ 9}$.
Câu 5:
\[A=3\sqrt{8}-\sqrt{50}-\sqrt{{{\left( \sqrt{2}-1 \right)}^{2}}}=6\sqrt{2}-5\sqrt{2}-\left( \sqrt{2}-1 \right)=1\]
b) $\text{B = }\frac{2}{\text{x - 1}}.\sqrt{\frac{{{\text{x}}^{\text{2}}}\text{ - 2x + 1}}{4{{\text{x}}^{\text{2}}}}}=\frac{2}{\text{x - 1}}\sqrt{\frac{{{\left( \text{x - 1} \right)}^{2}}}{{{2}^{2}}{{\text{x}}^{\text{2}}}}}=\frac{2}{\text{x - 1}}.\frac{\left| \text{x - 1} \right|}{2\left| \text{x} \right|}$
Vì 0 < x < 1 nên $\Rightarrow \text{B = }\frac{\text{- 2}\left( \text{x - 1} \right)}{\text{2x}\left( \text{x - 1} \right)}=-\frac{1}{\text{x}}$.
Câu 6: 1) Rút gọn
A = $\left[ \frac{\left( \text{1 - }\sqrt{\text{a}} \right)\text{ }\left( \text{1 + }\sqrt{\text{a}}\text{ + a} \right)}{\text{1 - }\sqrt{\text{a}}}\text{ + }\sqrt{\text{a}} \right]{{\left[ \frac{\text{1 - }\sqrt{\text{a}}}{\left( \text{1 - }\sqrt{\text{a}} \right)\text{ }\left( \text{1 + }\sqrt{\text{a}} \right)} \right]}^{2}}$
= $\left( \text{1 + 2}\sqrt{\text{a}}\text{ + a} \right).\frac{1}{{{\left( 1\text{ + }\sqrt{\text{a}} \right)}^{2}}}\text{ = }{{\left( 1\text{ + }\sqrt{\text{a}} \right)}^{2}}.\frac{1}{{{\left( 1\text{ +}\sqrt{\text{a}} \right)}^{2}}}\text{ = 1}\text{.}$
Câu 7: Rút gọn biểu thức
1) A = $\sqrt{20}\text{ - }\sqrt{45}\text{ + 3}\sqrt{18}\text{ + }\sqrt{72}$ = $\sqrt{5\text{ }\text{. 4}}\text{ - }\sqrt{\text{9 }\text{. 5}}\text{ + 3}\sqrt{9\text{ }\text{. 2}}\text{ + }\sqrt{\text{36 }\text{. 2}}$
= $2\sqrt{5}\text{ - 3}\sqrt{5}\text{ + 9}\sqrt{2}\text{ + 6}\sqrt{2}$ = 15$\sqrt{2}\text{ - }\sqrt{5}$
2) B = $\left( 1\text{ + }\frac{\text{a + }\sqrt{\text{a}}}{\sqrt{\text{a}}\text{ + 1}} \right)\left( \text{1 + }\frac{\text{a - }\sqrt{\text{a}}}{1\text{ - }\sqrt{\text{a}}} \right)$ với a ≥ 0, a ≠ 1
= $\left( 1\text{ + }\frac{\sqrt{\text{a}}\text{ (}\sqrt{\text{a}}\text{ + 1)}}{\sqrt{\text{a}}\text{ + 1}} \right)\left( \text{1 - }\frac{\sqrt{\text{a}}\text{ (}\sqrt{\text{a}}\text{ - 1)}}{\sqrt{\text{a}}\text{ - 1}} \right)$ = (1 + $\sqrt{\text{a}}$) (1 - $\sqrt{\text{a}}$) = 1 - a
Câu 8:
1) Điều kiện: a ≥ 0, a ≠ 1, a ≠ 2
Ta có: $\text{P = }\left[ \frac{\left( \sqrt{\text{a}}\text{ - 1} \right)\text{ }\left( \text{a + }\sqrt{\text{a}}\text{ + 1} \right)}{\sqrt{\text{a}}\text{ }\left( \sqrt{\text{a}}\text{ - 1} \right)}\text{ - }\frac{\left( \sqrt{\text{a}}\text{ + 1} \right)\text{ }\left( \text{a - }\sqrt{\text{a}}\text{ + 1} \right)}{\sqrt{\text{a}}\text{ }\left( \sqrt{\text{a}}\text{ + 1} \right)} \right]\text{ : }\frac{\text{a + 2}}{\text{a - 2}}$
$\text{ = }\frac{\text{a + }\sqrt{\text{a}}\text{ + 1 - a + }\sqrt{\text{a}}\text{ - 1}}{\sqrt{\text{a}}}\text{ : }\frac{\text{a + 2}}{\text{a - 2}}$ $\text{= }\frac{2\text{ (a - 2)}}{\text{a + 2}}$
Câu 9: 1) Ta có : $\text{P = }\frac{\sqrt{\text{x}}\text{ + 1}}{\sqrt{\text{x}}\text{ - 2}}\text{ + }\frac{2\sqrt{\text{x}}}{\sqrt{\text{x}}\text{ +}\text{2}}\text{ - }\frac{2\text{ + 5}\sqrt{\text{x}}}{\text{x - 4}}$
P = $\frac{(\sqrt{\text{x}}\text{+1) (}\sqrt{\text{x}}\text{ +}\text{2) + 2}\sqrt{\text{x}}\text{ (}\sqrt{\text{x}}\text{ - 2) - 2 - 5}\sqrt{\text{x}}}{(\sqrt{\text{x}}\text{ - 2) (}\sqrt{\text{x}}\text{ + 2)}}$ =
= $\frac{\text{x + 3}\sqrt{\text{x}}\text{ +}\text{2 + 2x - 4}\sqrt{\text{x}}\text{ - 2 - 5}\sqrt{\text{x}}}{(\sqrt{\text{x}}\text{ +}\text{2) (}\sqrt{\text{x}}\text{ - 2)}}$
= $\frac{\text{3x - 6}\sqrt{\text{x}}}{(\sqrt{\text{x}}\text{ + 2) (}\sqrt{\text{x}}\text{ - 2)}}\text{ = }\frac{3\sqrt{\text{x}}\text{ (}\sqrt{\text{x}}-2)}{(\sqrt{\text{x}}\text{ + 2) (}\sqrt{\text{x}}\text{ - 2)}}\text{ = }\frac{3\sqrt{\text{x}}}{\sqrt{\text{x}}\text{ +}\text{2}}$
Câu 10: a) M = $\left( \frac{\sqrt{\text{x}}}{\sqrt{\text{x}}\text{ - 1}}\text{ - }\frac{1}{\text{x - }\sqrt{\text{x}}} \right)\text{ : }\left( \frac{1}{\sqrt{\text{x}}\text{ + 1}}\text{ + }\frac{2}{\text{x - 1}} \right)$
= $\left[ \frac{\sqrt{\text{x}}}{\sqrt{\text{x}}\text{ - 1}}\text{ - }\frac{1}{\sqrt{\text{x}}\text{ (}\sqrt{\text{x}}\text{ - 1)}} \right]\text{ : }\left[ \frac{\sqrt{\text{x}}\text{ - 1}}{\left( \sqrt{\text{x}}\text{ - 1} \right)\text{ }\left( \sqrt{\text{x}}\text{ + 1} \right)}\text{ + }\frac{2}{\left( \sqrt{\text{x}}\text{ - 1} \right)\text{ }\left( \sqrt{\text{x}}\text{ +}\text{1} \right)} \right]$
= $\frac{\text{x - 1}}{\sqrt{\text{x}}\text{ }\left( \sqrt{\text{x}}\text{ - 1} \right)}\text{ : }\frac{\sqrt{\text{x}}\text{ + 1}}{\left( \sqrt{\text{x}}\text{ - 1} \right)\text{ }\left( \sqrt{\text{x}}\text{ +}\text{1} \right)}\text{ = }\frac{\text{x - 1}}{\sqrt{\text{x}}\left( \sqrt{\text{x}}\text{ - 1} \right)}\text{ }\text{. }\frac{\left( \sqrt{\text{x}}\text{ - 1} \right)\text{ }\left( \sqrt{\text{x}}\text{ + 1} \right)}{\sqrt{\text{x}}\text{ + 1}}$
= $\frac{\text{x - 1}}{\sqrt{\text{x}}}$.
Câu 11:
1) K = $\frac{\text{x}}{\sqrt{\text{x}}\text{ - 1}}\text{-}\frac{\sqrt{\text{x}}\text{(2}\sqrt{\text{x}}\text{ - 1)}}{\sqrt{\text{x}}\text{(}\sqrt{\text{x}}\text{ - 1)}}$ = $\frac{\text{x - 2}\sqrt{\text{x}}\text{ + 1}}{\sqrt{\text{x}}\text{ - 1}}\text{ = }\sqrt{\text{x}}\text{ - 1}$
2) Khi x = 4 + 2$\sqrt{3}$, ta có: K = $\sqrt{4+2\sqrt{3}}$- 1 = $\sqrt{{{\left( \sqrt{3}+1 \right)}^{2}}}-1=\sqrt{3}+1-1=\sqrt{3}$
Câu 12: Rút gọn biểu thức:
1) $\sqrt{45}+\sqrt{20}-\sqrt{5}$ = $\sqrt{{{3}^{2}}.5}+\sqrt{{{2}^{2}}.5}-\sqrt{5}$
= $3\sqrt{5}+2\sqrt{5}-\sqrt{5}$ = 4$\sqrt{5}$
2) $\frac{x+\sqrt{x}}{\sqrt{x}}+\frac{x-4}{\sqrt{x}+2}$= $\frac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}}+\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{\sqrt{x}+2}$
= $\sqrt{x}+1+\sqrt{x}-2$ = 2$\sqrt{x}-1$
Câu 13: a) A = $\frac{\sqrt{5}(\sqrt{5}+7)}{\sqrt{5}}+\frac{\sqrt{11}(\sqrt{11}+1)}{1+\sqrt{11}}=\sqrt{5}+7+\sqrt{11}.$
b) B = $\sqrt{5}.\frac{\sqrt{5}(\sqrt{5}+\sqrt{11})}{5}=\sqrt{5}+\sqrt{11}$.
Vậy \[A-B=\sqrt{5}+7+\sqrt{11}-\sqrt{5}-\sqrt{11}=7\]
Câu 14: a) $\text{A = }\frac{2(\sqrt{5}\text{ +}2)\text{ - 2(}\sqrt{5}\text{ - 2)}}{\left( \sqrt{5}\text{ - 2} \right)\text{ }\left( \sqrt{5}\text{ +}\text{2} \right)}\text{ = }\frac{2\sqrt{5}\text{ +}\text{4 - 2}\sqrt{5}\text{ + 4}}{{{\left( \sqrt{5} \right)}^{2}}\text{ - }{{\text{2}}^{2}}}\text{ = }\frac{8}{5\text{ - 4}}\text{ = 8}$.
b) Ta có:
$\text{B = }\frac{\text{x - 1}}{\sqrt{\text{x}}}\text{:}\frac{\left( \sqrt{\text{x}}\text{ - 1} \right)\left( \sqrt{\text{x}}\text{ + 1} \right)\text{ +}\text{1 - }\sqrt{\text{x}}}{\sqrt{\text{x}}\left( \sqrt{\text{x}}\text{ +}\text{1} \right)}\text{ = }\frac{\text{x - 1}}{\sqrt{\text{x}}}\cdot \frac{\sqrt{\text{x}}\left( \sqrt{\text{x}}\text{ +}\text{1} \right)}{\text{x - 1 + 1 - }\sqrt{\text{x}}}\text{ }\,\,\text{= }\frac{\left( \text{x - 1} \right)\left( \sqrt{\text{x}}\text{ +}\,\text{1} \right)}{\sqrt{\text{x}}\text{ }\left( \sqrt{\text{x}}\text{ - 1} \right)}=\frac{{{\left( \sqrt{\text{x}}\text{ +}\,\text{1} \right)}^{2}}}{\sqrt{\text{x}}\text{ }}$
Câu 15.
1) A = $(1-\sqrt{5})\cdot \frac{\sqrt{5}(1+\sqrt{5})}{2\sqrt{5}}=(1-\sqrt{5})\cdot \frac{(1+\sqrt{5})}{2}=\frac{1-5}{2}=-2$.
2) B = $\left( 1+\frac{\sqrt{x}\left( \sqrt{x}+1 \right)}{1+\sqrt{x}} \right)\left( 1+\frac{\sqrt{x}\left( \sqrt{x}-1 \right)}{1-\sqrt{x}} \right)=\left( 1+\sqrt{x} \right)\left( 1-\sqrt{x} \right)=1-x$.
Câu 16.
1) Ta có A = $\left( \frac{x-1}{\sqrt{x}\left( \sqrt{x}-1 \right)} \right):\left( \frac{\sqrt{x}+1}{x-1} \right)$ = $\frac{\sqrt{x}+1}{\sqrt{x}}.\frac{x-1}{\sqrt{x}+1}=\frac{x-1}{\sqrt{x}}$.
Câu 17:
1) $\text{ P = }\left( \frac{\text{1}}{\text{x + }\sqrt{\text{x}}}-\frac{1}{\sqrt{\text{x}}+1} \right):\frac{\sqrt{\text{x}}}{\text{x + 2}\sqrt{\text{x}}+1}$ $=\left( \frac{1}{\sqrt{\text{x}}\left( \sqrt{\text{x}}+1 \right)}-\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}\left( \sqrt{\text{x}}+1 \right)} \right).\frac{{{\left( \sqrt{\text{x}}+1 \right)}^{2}}}{\sqrt{\text{x}}}$
$=\frac{1-\sqrt{\text{x}}}{\sqrt{\text{x}}\left( \sqrt{\text{x}}+1 \right)}.\frac{{{\left( \sqrt{\text{x}}+1 \right)}^{2}}}{\sqrt{\text{x}}}=\frac{\left( 1-\sqrt{\text{x}} \right)\left( \sqrt{\text{x}}+1 \right)}{\sqrt{\text{x}}.\sqrt{\text{x}}}=\frac{\text{1 - x }}{\text{x}}$.
2) Với x > 0 thì $\frac{\text{1 - x}}{\text{x}}>\frac{1}{2}\Leftrightarrow 2\left( \text{1 - x} \right)>\text{x}\Leftrightarrow \text{-3x+}2>0\Leftrightarrow x<\frac{2}{3}$
Câu 18:
1) A = $\frac{1}{2}\sqrt{4.5}-\sqrt{16.5}+\frac{2}{3}\sqrt{9.5}$ = $\sqrt{5}-4\sqrt{5}+2\sqrt{5}$ = $-\sqrt{5}$.
2) $\text{B = }\left( 2+\frac{5-\sqrt{5}}{\sqrt{5}-1} \right).\left( 2-\frac{5+\sqrt{5}}{\sqrt{5}+1} \right)$
$=\left( 2+\frac{\sqrt{5}\left( \sqrt{5}-1 \right)}{\sqrt{5}-1} \right)\left( 2-\frac{\sqrt{5}\left( \sqrt{5}+1 \right)}{\sqrt{5}+1} \right)=\left( 2+\sqrt{5} \right)\left( 2-\sqrt{5} \right)=-1$
Câu 19.
1)$\text{ A =}\left( \frac{\sqrt{\text{a}}}{\sqrt{\text{a}}+1}-\frac{\sqrt{\text{a}}}{\sqrt{\text{a}}\text{(}\sqrt{\text{a}}\text{ + 1)}} \right):\frac{\sqrt{\text{a}}-1}{(\sqrt{\text{a}}\text{ - 1)(}\sqrt{\text{a}}+1)}$ $\text{ }=\left( \frac{\sqrt{\text{a}}}{\sqrt{\text{a}}+1}-\frac{\text{1}}{\sqrt{\text{a}}\text{ + 1}} \right).\left( \sqrt{\text{a}}+1 \right)=\sqrt{\text{a}}-1$
2) A < 0 .png)
Câu 20:
a) P = \[\left( \frac{1}{\sqrt{a}-3}+\frac{1}{\sqrt{a}+3} \right).\left( 1-\frac{3}{\sqrt{a}} \right)=\frac{\sqrt{a}+3+\sqrt{a}-3}{\left( \sqrt{a}-3 \right)\left( \sqrt{a}+3 \right)}.\frac{\sqrt{a}-3}{\sqrt{a}}\].
= \[\frac{2\sqrt{a}.(\sqrt{a}-3)}{(\sqrt{a}-3)(\sqrt{a}+3).\sqrt{a}}=\frac{2}{\sqrt{a}+3}\]. Vậy P = \[\frac{2}{\sqrt{a}+3}\].
b) Ta có: \[\frac{2}{\sqrt{a}+3}\] > \[\frac{1}{2}\] $\Leftrightarrow $ \[\sqrt{a}\] + 3 < 4 $\Leftrightarrow $\[\sqrt{a}\] < 1
Vậy P > \[\frac{1}{2}\] khi và chỉ khi 0 < a < 1.
Câu 21. 1) Ta có $9\sqrt{a}-\sqrt{25a}+\sqrt{4{{a}^{3}}}=9\sqrt{a}-5\sqrt{a}+2a\sqrt{a}$ $=2\sqrt{a}(a+2)$ và ${{a}^{2}}+2a=a(a+2)$
nên P = \[\frac{2\sqrt{a}\left( a+2 \right)}{a\left( a+2 \right)}=\frac{2}{\sqrt{a}}\] .
Câu 22: Tính
a) A = \[\sqrt{20}-3\sqrt{18}-\sqrt{45}+\sqrt{72}=\sqrt{4.5}-3\sqrt{9.2}-\sqrt{9.5}+\sqrt{36.2}\] =
= $2\sqrt{5}-9\sqrt{2}-3\sqrt{5}+6\sqrt{2}=-3\sqrt{2}-\sqrt{5}$.
b) B = \[\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\]
\[\sqrt{2}B=\sqrt[{}]{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}=\sqrt{{{(\sqrt{7}+1)}^{2}}}+\sqrt{{{(\sqrt{7}-1)}^{2}}}=\sqrt{7}+1+\left| \sqrt{7}-1 \right|\]
\[\sqrt{2}B=2\sqrt{7}\Leftrightarrow B=\sqrt{14}\]
c) C = \[\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\] với x > 1
C = \[\sqrt{{{(\sqrt{x-1}+1)}^{2}}}+\sqrt{{{(\sqrt{x-1}-1)}^{2}}}=\sqrt{x-1}+1+\left| \sqrt{x-1}-1 \right|\]
+) Nếu x > 2 thì C = \[\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\]
+) Nếu x < 2, thì C = \[\sqrt{x-1}+1+1-\sqrt{x-1}=2\].
Câu 23: 1) P = $(\sqrt{7}+\sqrt{3}-2)(\sqrt{7}-\sqrt{3}+2)=\text{ }\!\![\!\!\text{ }\sqrt{7}+(\sqrt{3}-2)\text{ }\!\!]\!\!\text{ }\,\text{ }\!\![\!\!\text{ }\sqrt{7}-(\sqrt{3}-2)\text{ }\!\!]\!\!\text{ }$
= \[{{(\sqrt{7})}^{2}}-(\sqrt{3}-2){{)}^{2}}=\,7-(3-4\sqrt{3}+4)=4\sqrt{3}\].
Câu 24: a) A = \[\left( \frac{a+1-2\sqrt{a}}{a+1} \right):\left[ \frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{\sqrt{a}(a+1)+(a+1)} \right]\]
= \[\frac{{{(\sqrt{a}-1)}^{2}}}{a+1}:\left[ \frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{(a+1)(\sqrt{a}+1)} \right]=\frac{{{(\sqrt{a}-1)}^{2}}}{a+1}:\frac{{{(\sqrt{a}-1)}^{2}}}{(\sqrt{a}+1)(a+1)}\].
= \[\frac{{{(\sqrt{a}-1)}^{2}}}{a+1}.\frac{(a+1)(\sqrt{a}+1)}{{{(\sqrt{a}-1)}^{2}}}=\sqrt{a}+1\].
b) a = 2011 - 2 \[\sqrt{2010}={{(\sqrt{2010}-1)}^{2}}\Rightarrow \sqrt{a}=\sqrt{2010}-1\]
Vậy A = \[\sqrt{2010}\].
Câu 25: P = \[\left| \sqrt{a}-1+1 \right|+\left| \sqrt{a}-1-1 \right|\]
Nếu a> 2 => \[\sqrt{a}-1-1\ge 0\Rightarrow P=2\sqrt{a}-1\]
Nếu 1< a < 2 => \[\sqrt{a}-1-1\] < 0 => P = 2
Câu 26: ĐKXĐ: x > 0; x \[\ne \] 1.
1) Q = \[\frac{{{(x-1)}^{2}}}{4x}.\frac{{{(\sqrt{x}+1)}^{2}}-{{(\sqrt{x}-1)}^{2}}}{x-1}=\frac{{{(x-1)}^{2}}.4\sqrt{x}}{4x.(x-1)}=\frac{x-1}{\sqrt{x}}\].
2) Q = - 3\[\sqrt{x}-3\] => 4x + 3\[\sqrt{x}\] - 1 = 0.png)
(thỏa mãn)
Câu 27: A = \[\frac{\sqrt{{{(x+3)}^{2}}}}{x+3}=\frac{\left| x+3 \right|}{x+3}\] = .png)
Câu 28: a) P = \[\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}+1}{\sqrt{a}-3}+\frac{-7\sqrt{a}-3}{(\sqrt{a}-3)(\sqrt{a}+3)}\]
= \[\frac{2\sqrt{a}(\sqrt{a}-3)+(\sqrt{a}+1)(\sqrt{a}+3)-7\sqrt{a}-3}{(\sqrt{a}-3)(\sqrt{a}+3)}=\frac{2a-6\sqrt{a}+a+4\sqrt{a}+3-7\sqrt{a}-3}{(\sqrt{a}-3)(\sqrt{a}+3)}\]
= \[\frac{3a-9\sqrt{a}}{(\sqrt{a}-3)(\sqrt{a}+3)}=\frac{3\sqrt{a}(\sqrt{a}-3)}{(\sqrt{a}-3)(\sqrt{a}+3)}=\frac{3\sqrt{a}}{\sqrt{a}+3}\]
Vậy P = \[\frac{3\sqrt{a}}{\sqrt{a}+3}\].
b) P < 1 $\Leftrightarrow $\[\frac{3\sqrt{a}}{\sqrt{a}+3}<1\Leftrightarrow 3\sqrt{a}<\sqrt{a}+3\Leftrightarrow \sqrt{a}<\frac{3}{2}\Leftrightarrow 0\le a<\frac{9}{4}\].
Câu 29: M = \[\frac{\sqrt{x}(\sqrt{{{x}^{3}}}-1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{{{x}^{3}}}+1)}{x-\sqrt{x}+1}\]+ x + 1
= \[\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}+1)(x-\sqrt{x}+1)}{x-\sqrt{x}+1}+x+1\]
= x - \[\sqrt{x}\]- x - \[\sqrt{x}\] + x + 1 = x - 2\[\sqrt{x}\] + 1 = (\[\sqrt{x}\] - 1)2
Câu 30:
a) Ta có x2 + \[\sqrt{x}=\sqrt{x}(\sqrt{{{x}^{3}}}+1)=\sqrt{x}(\sqrt{x}+1)(x-\sqrt{x}+1)\]
nên P = \[\frac{\sqrt{x}(\sqrt{x}+1)(x-\sqrt{x}+1)}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}\]
= \[\sqrt{x}(\sqrt{x}+1)+1-2\sqrt{x}-1=x-\sqrt{x}\]. Vậy P = $x-\sqrt{x}$.
b) P = 0 $\Leftrightarrow $ x - \[\sqrt{x}\] = 0 $\Leftrightarrow $\[\sqrt{x}\](\[\sqrt{x}\] - 1) = 0 $\Leftrightarrow $ x = 0 (loại) ; x = 1 (t/m)
Vậy x = 1 thì P = 0
Câu 31:
1) Tính: $\sqrt{\text{48}}\text{ - 2}\sqrt{\text{75}}\text{ + }\sqrt{\text{108}}$= $\sqrt{\text{16 }\text{. 3}}\text{ - 2}\sqrt{\text{25 }\text{. 3}}\text{ + }\sqrt{\text{36 }\text{. 3}}$
= \[\text{4}\sqrt{\text{3}}\text{ - 10}\sqrt{\text{3}}\text{ + 6}\sqrt{\text{3}}\text{ = 0}\]
2) Rút gọn biểu thức: P = $\left( \frac{\text{1}}{\text{1 - }\sqrt{\text{x}}}\text{ - }\frac{\text{1}}{\text{1 + }\sqrt{\text{x}}} \right)\text{ }.\text{ }\left( \text{1 - }\frac{\text{1}}{\sqrt{\text{x}}} \right)$
= $\left( \frac{\text{1 + }\sqrt{\text{x}}\text{ - 1 +}\sqrt{\text{x}}}{\text{1- x}} \right)\left( \frac{\sqrt{\text{x}}\text{ - 1}}{\sqrt{\text{x}}} \right)$ =$\frac{\text{2}\sqrt{\text{x}}}{\text{1- x}}\text{ }\text{. }\frac{\sqrt{\text{x}}\text{ - 1}}{\sqrt{\text{x}}}$ = $\frac{\text{- 2}}{\text{1 + }\sqrt{\text{x}}}$
