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