đk : x >= 2
\(3\sqrt{x-2}+2x=\sqrt{x+6}+6\)
\(\Leftrightarrow3\sqrt{x-2}-3+2x-6-\left(\sqrt{x+6}-3\right)=0\)
\(\Leftrightarrow\frac{9\left(x-2\right)-9}{3\sqrt{x-2}+3}+2\left(x-3\right)-\frac{x+6-9}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow\frac{9x-27}{3\sqrt{x-2}+3}+2\left(x-3\right)-\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{9}{3\sqrt{x-2}+3}+2-\frac{1}{\sqrt{x+6}+3}\ne0\right)=0\Leftrightarrow x=3\)(tmđk)