\(M=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\left(\sqrt{x}+1\right)+x-2}{x\left(\sqrt{x}+1\right)}\)
\(M=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x\left(\sqrt{x}+1\right)}{x+2\sqrt{x}}\)
\(M=\dfrac{x}{\sqrt{x}-1}\)
b) \(M=-\dfrac{1}{2}\Leftrightarrow\dfrac{x}{\sqrt{x}-1}=-\dfrac{1}{2}\Leftrightarrow\sqrt{x}-1=-2x\Leftrightarrow2x+\sqrt{x}-1=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\Leftrightarrow2\sqrt{x}-1=0\Leftrightarrow x=\dfrac{1}{4}\)
c) \(M>1\Leftrightarrow\dfrac{x}{\sqrt{x}-1}>1\)
\(\Leftrightarrow\dfrac{x}{\sqrt{x}-1}-1>0\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)
Ta có: \(x-\sqrt{x}+1=x-2\cdot\dfrac{1}{2}\cdot\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{4}\right)^2+\dfrac{3}{4}>0\)
\(\Rightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)
