ĐKXĐ : x >= 0, x khác 9.
Ta có \(A=\left(\dfrac{\sqrt{x}}{x-9}+\dfrac{2}{\sqrt{x}+3}+\dfrac{3}{3-\sqrt{x}}\right):\left(\sqrt{x}-3+\dfrac{12-x}{\sqrt{x}+3}\right)=\left(\dfrac{\sqrt{x}}{x-9}+\dfrac{6-2\sqrt{x}-3\sqrt{x}-9}{x-9}\right):\dfrac{x-9+12-x}{\sqrt{x}+3}=\dfrac{-3}{x-9}:\dfrac{3}{\sqrt{x}+3}=\dfrac{-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}+3}{3}=\dfrac{-3}{\sqrt{x}-3}\)