Question

Answer 1

\(=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}}=\dfrac{2}{\sqrt{x}-2}\)