Câu hỏi của Mai Anh Nguyễn

Question

Cho biểu thức: Q= $\left(\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}\right)+\frac{3-\sqrt{x}}{x-1}$

a) Tính giá trị của Q

b) Tìm x sao cho Q=-1

Ngân Vũ Thị · Accepted Answer

https://i.imgur.com/k8yqeDi.jpgCâu trả lời: https://i.imgur.com/k8yqeDi.jpg

Nguyễn Thành Trương · Answer

ĐK: $x\ge0,x\ne1$

$\begin{array}{l}
a)Q = \left( {\dfrac{{\sqrt x }}{{1 - \sqrt x }} + \dfrac{{\sqrt x }}{{1 + \sqrt x }}} \right) + \dfrac{{3 - \sqrt x }}{{x - 1}}\
Q = \dfrac{{\left( {1 + \sqrt x } \right)\sqrt x  + \left( {1 - \sqrt x } \right)\sqrt x }}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} + \dfrac{{3 - \sqrt x }}{{x - 1}}\
Q = \dfrac{{\sqrt x  + x + \sqrt x  - x}}{{1 - x}} + \dfrac{{3 - \sqrt x }}{{x - 1}}\
Q = \dfrac{{2\sqrt x }}{{ - \left( {x - 1} \right)}} + \dfrac{{3 - \sqrt x }}{{x - 1}}\
Q = \dfrac{{ - 2\sqrt x  + 3 - \sqrt x }}{{x - 1}}\
Q = \dfrac{{ - 3\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\
Q = \dfrac{{ - 3}}{{\sqrt x  + 1}}
\end{array}
$

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\begin{array}{l}
b)Q =  - 1 \Rightarrow \dfrac{{ - 3}}{{\sqrt x  + 1}} =  - 1\
 \Leftrightarrow  - \left( {\sqrt x  + 1} \right) =  - 3\
 \Leftrightarrow  - \sqrt x  - 1 =  - 3\
 \Leftrightarrow  - \sqrt x  =  - 3 + 1\
 \Leftrightarrow  - \sqrt x  =  - 2\
 \Leftrightarrow {\left( { - \sqrt x } \right)^2} = {\left( { - 2} \right)^2}\
 \Leftrightarrow x = 4
\end{array}$Câu trả lời: ĐK: $x\ge0,x\ne1$

$\begin{array}{l}
a)Q = \left( {\dfrac{{\sqrt x }}{{1 - \sqrt x }} + \dfrac{{\sqrt x }}{{1 + \sqrt x }}} \right) + \dfrac{{3 - \sqrt x }}{{x - 1}}\
Q = \dfrac{{\left( {1 + \sqrt x } \right)\sqrt x  + \left( {1 - \sqrt x } \right)\sqrt x }}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} + \dfrac{{3 - \sqrt x }}{{x - 1}}\
Q = \dfrac{{\sqrt x  + x + \sqrt x  - x}}{{1 - x}} + \dfrac{{3 - \sqrt x }}{{x - 1}}\
Q = \dfrac{{2\sqrt x }}{{ - \left( {x - 1} \right)}} + \dfrac{{3 - \sqrt x }}{{x - 1}}\
Q = \dfrac{{ - 2\sqrt x  + 3 - \sqrt x }}{{x - 1}}\
Q = \dfrac{{ - 3\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\
Q = \dfrac{{ - 3}}{{\sqrt x  + 1}}
\end{array}
$

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\begin{array}{l}
b)Q =  - 1 \Rightarrow \dfrac{{ - 3}}{{\sqrt x  + 1}} =  - 1\
 \Leftrightarrow  - \left( {\sqrt x  + 1} \right) =  - 3\
 \Leftrightarrow  - \sqrt x  - 1 =  - 3\
 \Leftrightarrow  - \sqrt x  =  - 3 + 1\
 \Leftrightarrow  - \sqrt x  =  - 2\
 \Leftrightarrow {\left( { - \sqrt x } \right)^2} = {\left( { - 2} \right)^2}\
 \Leftrightarrow x = 4
\end{array}$

Võ Hồng Phúc · Answer

$a.$

$Q=\left(\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}\right)+\frac{3-\sqrt{x}}{x-1}$ $ĐKXĐ:x\ge0;x\ne1$

$=\left(\frac{\sqrt{x}.\left(1+\sqrt{x}\right)}{1-x}+\frac{\sqrt{x}.\left(1-\sqrt{x}\right)}{1-x}\right)+\frac{\sqrt{x}-3}{1-x}$

$=\frac{\sqrt{x}+x+\sqrt{x}-x+\sqrt{x}-3}{1-x}$

$=\frac{3\sqrt{x}-3}{1-x}=-\frac{3\left(\sqrt{x}-1\right)}{x-1}=-\frac{3}{\sqrt{x}+1}$

$b.$

$Q=-1\Leftrightarrow-\frac{3}{\sqrt{x}+1}=-1$

$\Leftrightarrow-\sqrt{x}-1=-3\Leftrightarrow\sqrt{x}=2$

$\Leftrightarrow x=4$

$\text{Vậy }Q=-1\Leftrightarrow x=4$Câu trả lời: $a.$