a) ĐKXĐ: x\(\ge0,x\ne1\)
A = \(\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\frac{\sqrt{x}-1}{2}\)
= \(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x +\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
= \(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
= \(\frac{2}{x+\sqrt{x}+1}\)
b) Ta có x\(\ge0,x\ne1\) =>\(x+\sqrt{x}+1>0\Rightarrow\frac{2}{x+\sqrt{x}+1}>0\)
=> A>0 (1)
Mặt khác \(x\ge0,x\ne1\Rightarrow x+\sqrt{x}+1\ge1\)
\(\Rightarrow\frac{2}{x+\sqrt{x}+1}\le2\) \(\Rightarrow A\ge2\) (2)
Từ (1) và (2) => \(0< A\le2\)
