Đặt x\(^2\)-2x=t

Do đó ta có: t\(^2\)-3t+2=0

\(\Leftrightarrow\) t\(^2\)-t-2t+2=0

\(\Leftrightarrow\) t(t-1)-2(t-1)=0

\(\Leftrightarrow\) (t-1)(t-2)=0

\(\Leftrightarrow\)\(\left[\begin{matrix}t-1=0\\t-2=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[\begin{matrix}t=1\\t=2\end{matrix}\right.\)

-Với t=1\(\Rightarrow\)x\(^2\)-2x=1\(\Rightarrow\)x\(^2\)-2x-1=0\(\Leftrightarrow\)\(\left[\begin{matrix}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{matrix}\right.\)

-Với t=2\(\Rightarrow\)x\(^2\)-2x=2\(\Rightarrow\)x\(^2\)-2x-2=0\(\Leftrightarrow\)\(\left[\begin{matrix}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{matrix}\right.\)