Lời giải:
ĐKXĐ: $x>0; x\neq 1$
\(\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{\sqrt{x}-x}\right):\left(\frac{1}{\sqrt{x}+1}-\frac{2}{1-\sqrt{x}}\right)=\left(\frac{x}{x-\sqrt{x}}-\frac{1}{x-\sqrt{x}}\right):\frac{1-\sqrt{x}-2\sqrt{x}-2}{(\sqrt{x}+1)(1-\sqrt{x})}\)
\(=\frac{x-1}{x-\sqrt{x}}:\frac{-(1+3\sqrt{x})}{1-x}=\frac{x-1}{x-\sqrt{x}}.\frac{x-1}{3\sqrt{x}+1}=\frac{(x-1)^2}{(x-\sqrt{x})(3\sqrt{x}+1)}\)