\(Đkxđ:x\ge3+\sqrt{12}\) \(\sqrt{x^2+6x-3}=x-3\)
\(\Leftrightarrow\left(\sqrt{x^2+6x-3}\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow x^2+6x-3=x^2-6x+9\)
\(\Leftrightarrow x^2+6x-3-x^2+6x-9=0\)
\(\Leftrightarrow12x-12=0\)
\(\Leftrightarrow12x=12\)
\(\Leftrightarrow x=1\)