Đk: x >/ 2
\(\sqrt{x^2-3x+2}+3=3\sqrt{x-1}+\sqrt{x-2}\)
\(\Leftrightarrow x^2-3x+2+9+6\sqrt{x^2-3x+2}=9x-9+x-2+6\sqrt{x^2-3x+2}\)
\(\Leftrightarrow x^2-13x+22=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(N\right)\\x=11\left(N\right)\end{matrix}\right.\)
KL: x=2, x=11