Ta có: \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)
\(\Leftrightarrow\left|x-3\right|=\sqrt{3}+1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=\sqrt{3}+1\left(x\ge3\right)\\x-3=-\sqrt{3}-1\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{3}\left(nhận\right)\\x=2-\sqrt{3}\left(nhận\right)\end{matrix}\right.\)