a) \(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}+x}\right)\left(ĐK:x>0;x\ne1\right)\)
\(=\left[\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{2}{x}-\frac{2-x}{x\left(\sqrt{x}+1\right)}\right]\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{x\left(\sqrt{x}+1\right)}{2\sqrt{x}+2-2+x}\)
\(=\frac{x+2\sqrt{x}}{\sqrt{x}-1}\cdot\frac{x}{2\sqrt{x}+x}=\frac{x}{\sqrt{x}-1}\)
b)Để P>2
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}>2\)
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-2>0\)
\(\Leftrightarrow\frac{x-2\sqrt{x}+2}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-1\right)^2+1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow x>1\left(tm\right)\)
Vậy x>1 thì P>2