ĐKXĐ : \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-1\ne0\\x-\sqrt{x}\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne1\\\sqrt{x}\left(\sqrt{x}-1\right)\ne0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có : \(A=\frac{x}{\sqrt{x}-1}-\frac{2x-\sqrt{x}}{x-\sqrt{x}}\)
=> \(A=\frac{x}{\sqrt{x}-1}-\frac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
=> \(A=\frac{x\sqrt{x}-\left(2x-\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
=> \(A=\frac{x\sqrt{x}-2x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
=> \(A=\frac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
=> \(A=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\)
=> \(A=\sqrt{x}-1\)
- Thay \(x=3+2\sqrt{2}\) vào A ta được :
\(A=\sqrt{3+2\sqrt{2}}-1\)
=> \(A=\sqrt{\left(1+\sqrt{2}\right)^2}-1\)
=> \(A=1+\sqrt{2}-1\)
=> \(A=\sqrt{2}\)
