\(đkx>0;x\ne9\\ A=\left(\dfrac{\sqrt{x}+3+\sqrt{x}-3}{x-9}\right)\left(\dfrac{\sqrt{x}-3}{\sqrt{x}}\right)\\ A=\dfrac{2\sqrt{x}}{x-9}.\dfrac{\sqrt{x}-3}{\sqrt{x}}=\dfrac{2}{\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{2}{\sqrt{x}+3}\)