Đặt \(\sqrt{x^2-6x+12}=t\ge\sqrt{3}\)
\(\Rightarrow f\left(t\right)=6t+8-t^2\)
\(f'\left(t\right)=-2t+6=0\Rightarrow t=3\)
\(\Rightarrow M=f\left(3\right)\)
Dấu "=" xảy ra khi \(t=3\)
\(\Rightarrow\sqrt{x^2-6x+12}=3\)
\(\Rightarrow x^2-6x+3=0\)
Theo Viet ta có: \(x_1x_2=3\)