ĐKXĐ: \(x\ge3\)
Ta có: \(\frac{2}{3}\sqrt{9x-27}+\sqrt{x-3}=6+\sqrt{4x-12}\)
\(\Rightarrow\frac{2}{3}.3\sqrt{x-3}+\sqrt{x-3}=6+2\sqrt{x-3}\)
\(\Rightarrow2\sqrt{x-3}+\sqrt{x-3}-2\sqrt{x-3}=6\)
\(\Rightarrow\sqrt{x-3}=6\Rightarrow x-3=36\Rightarrow x=39\)
Vậy x = 39