Đặt \(\sqrt{x^2-2x+3}=\sqrt{\left(x-1\right)^2+2}=t\ge\sqrt{2}\)
\(\Rightarrow2x-x^2=3-t^2\)
\(\Rightarrow y=-t^2+4t+3=-\left(t-2\right)^2+7\le7\)
\(y_{max}=7\) khi \(t=2\Leftrightarrow\sqrt{x^2-2x+3}=2\)
\(\Leftrightarrow x^2-2x-1=0\Rightarrow x_1x_2=-1\)