\(\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{x\sqrt{x}-9\sqrt{x}}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{3\sqrt{x}-3}{x+3\sqrt{x}}\right)\left(x>0;x\ne9\right)\\ =\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+3\right)}\right]\\ =\dfrac{x-3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-3\sqrt{x}+3}\\ =\dfrac{x-3\sqrt{x}+1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-3x+3\right)}\)
