\(A=\dfrac{3}{\sqrt{x}+2}+\dfrac{3}{\sqrt{x}-2}\left(dkxd:x\ne4,x\ge0\right)\)
\(=\dfrac{3\left(\sqrt{x}-2\right)+3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6+3\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{6\sqrt{x}}{x-4}\)
Vậy \(A=\dfrac{6\sqrt{x}}{x-4}\) với \(x\ne4,x\ge0\)