C = $\frac{\sqrt{x}}{\sqrt{x}-2}$ = $1+\frac{2}{\sqrt{x}-2}$

 ta biết |C|>C khi C < 0 nên

$\frac{2}{\sqrt{x}-2}+1<0$

hay  $\frac{2}{\sqrt{x}-2}<-1$

2<$-1(\sqrt{x}-2)=2-\sqrt{x}$

hay $\sqrt{x}>0$

suy ra x>0