a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne0\\\sqrt{x}+1\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có : \(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}+x}\right)\)
=> \(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}-\frac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)
=> \(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\right)\)
=> \(P=\left(\frac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\right)\)
=> \(P=\frac{x\left(x+2\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)}\)
=> \(P=\frac{x}{\sqrt{x}-1}\)
b, Ta có : P > 2
=> \(\frac{x}{\sqrt{x}-1}>2\)
=> \(x>2\sqrt{x}-2\)
=> \(x-2\sqrt{x}+2>0\)
=> \(\left(\sqrt{x}-1\right)^2+1>0\)
