Ta có: \(A=\left(\frac{\sqrt{x}+2}{x-1}-\frac{\sqrt{x}-2}{x-2\sqrt{x}+1}\right):\frac{4x}{\left(x-1\right)^2}\)
\(=\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)}\right)\cdot\frac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{4x}\)
\(=\frac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{4x}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}-2-x+\sqrt{x}+2\right)}{4x}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{2\sqrt{x}\cdot2\sqrt{x}}=\frac{\sqrt{x}+1}{2\sqrt{x}}\)
