A=\(\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\) (đk: \(x\ge0,x\ne1\))
= \(\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-1}{2}\)
=\(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
=\(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
= \(\frac{2.\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{2}{x+\sqrt{x}+1}\)
b,Có A= \(\frac{2}{x+\sqrt{x}+1}=\frac{2}{x+2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}+\frac{3}{4}}=\frac{2}{\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}}\)
Có: \(\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\) vs mọi x khác 1
=> \(\frac{2}{\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}}>0\) với mọi x khác 1
<=> A>0 vói mọi x khác 1
