a) \(\left(\text{ }\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{1+\sqrt{x}}+\frac{2}{x-1}\right)\)
\(=\left(\frac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
b) ĐK : \(x>1\)
Ta có : \(A=\left(\sqrt{x}-1\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
Vì \(\frac{\sqrt{x}+1}{\sqrt{x}}>0\forall x\)
\(\Rightarrow\sqrt{x}-1>0\)
\(\Leftrightarrow\sqrt{x}>1\)
\(\Leftrightarrow x>1\)
Vậy với mọi x > 1 thì A > 0
