a) Ta có: \(A=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\dfrac{x^2-2x+1}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{1}\cdot\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{-2\sqrt{x}\cdot\left(\sqrt{x}-1\right)}{2}\)
\(=\sqrt{x}\left(1-\sqrt{x}\right)\)
b) Để A>0 thì \(\sqrt{x}\left(1-\sqrt{x}\right)>0\)
\(\Leftrightarrow1-\sqrt{x}>0\)
hay x<1
Kết hợp ĐKXĐ, ta được: 0<x<1
