A= ($\frac{1}{√x -1}$ + $\frac{1}{√x +1}$) : $\frac{√x}{x-1}$ (x>0 ,x $\neq$ 1)

A= ( $\frac{1}{\sqrt[]{x}-1}$ +$\frac{1}{\sqrt[]{x}+1}$): $\frac{\sqrt[]{x}}{x-1}$

= ($\frac{\sqrt[]{x}+1}{x-1}$ +$\frac{\sqrt[]{x}-1}{x-1}$): $\frac{\sqrt[]{x}}{x-1}$ 

= $\frac{2\sqrt[]{x}}{x-1}$ . $\frac{x-1}{\sqrt[]{x}}$ 

= 2

Vậy A = 2 với x>0,x$\neq$ 1