Question

Answer 1

Đặt \(\sqrt{x^2-3x+3}=t>0\Rightarrow x^2-3x+6=t^2+3\)

Pt trở thành: \(t+\sqrt{t^2+3}=3\)

\(\Leftrightarrow\sqrt{t^2+3}=3-t\)

\(\Leftrightarrow\left\{{}\begin{matrix}3-t\ge0\\t^2+3=\left(3-t\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\le3\\-6=-6t\end{matrix}\right.\) \(\Rightarrow t=1\)

\(\Rightarrow\sqrt{x^2-3x+3}=1\)

\(\Leftrightarrow x^2-3x+2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)