đặt t = x2 - 2x
t2 - 3t + 2 = 0 pt có dạng a+b+c =0
t1 = 1; t2 =c/a = 2/1=2
thay vào tinh dc x
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\(\Leftrightarrow\)\(\left[\begin{matrix}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{matrix}\right.\)
