a) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(\sqrt{x}-1\right)=\dfrac{x-1}{\sqrt{x}}\)
b) \(A=2\Rightarrow\dfrac{x-1}{\sqrt{x}}=2\Rightarrow x-1=2\sqrt{x}\Rightarrow x-2\sqrt{x}-1=0\)
\(\Rightarrow x-2\sqrt{x}+1=2\Rightarrow\left(\sqrt{x}-1\right)^2=2\Rightarrow\left[{}\begin{matrix}\sqrt{x}-1=\sqrt{2}\\\sqrt{x}-1=-\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{2}+1\\\sqrt{x}=1-\sqrt{2}\left(l\right)\end{matrix}\right.\Rightarrow x=\left(\sqrt{2}+1\right)^2=3+2\sqrt{2}\)
c) \(A< 0\Rightarrow\dfrac{x-1}{\sqrt{x}}< 0\) mà \(\sqrt{x}>0\Rightarrow x-1< 0\Rightarrow x< 1\Rightarrow0< x< 1\)
a) Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{\sqrt{x}+1}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
b) Để A=2 thì \(x-1=2\sqrt{x}\)
\(\Leftrightarrow x-2\sqrt{x}+1=2\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\sqrt{x}-1=\sqrt{2}\)
\(\Leftrightarrow\sqrt{x}=\sqrt{2}+1\)
hay \(x=3+2\sqrt{2}\)
c) Để A<0 thì x-1<0
hay x<1
Kết hợp ĐKXĐ, ta được: 0<x<1
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