ĐK:x>0,x\(\ne\)9
\(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right)\div\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right)\div\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\div\left[\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\div\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\dfrac{-3\left(\sqrt{x}+3\right).\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)\left(2\sqrt{x}+4\right)}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
