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