\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\right):\dfrac{1}{x+6\sqrt{x}+9}\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\left(x+6\sqrt{x}+9\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\left(\sqrt{x}+3\right)^2\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\left(\sqrt{x}+3\right)^2=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\left(\sqrt{x}+3\right)^2\)
\(=3\left(\sqrt{x}+3\right)=3\sqrt{x}+9\)
Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\right):\dfrac{1}{x+6\sqrt{x}+9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{1}\)
\(=\dfrac{3\sqrt{x}-9}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}+3}{1}\)
\(=3\sqrt{x}+9\)