a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\dfrac{2}{\sqrt{x}+2}\)