a) M \(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\right)\)
= \(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x\left(\sqrt{x}+1\right)}{x+2\sqrt{x}}\)
= \(\dfrac{x}{\sqrt{x}-1}\)
b) Ta có \(\dfrac{x}{\sqrt{x}-1}\)= \(\dfrac{-1}{2}\)
\(\Leftrightarrow2x=1-\sqrt{x}\)
\(\Leftrightarrow2x+\sqrt{x}-1=0\)
\(\Leftrightarrow2x+2\sqrt{x}-\sqrt{x}-1\) \(=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)
Vì:
\(\sqrt{x}+1>0\Rightarrow2\sqrt{x}-1=0\\ \Rightarrow x=\dfrac{1}{4}\left(tm\right)\)
a) Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b) Để \(M=-\dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}}{\sqrt{x}-1}=\dfrac{-1}{2}\)
\(\Leftrightarrow2\sqrt{x}=-\sqrt{x}+1\)
\(\Leftrightarrow3\sqrt{x}=1\)
hay \(x=\dfrac{1}{9}\)(thỏa ĐK)
