Question

Giải phương trình bằng cách đặt ẩn

\(\left(x^2-2x\right)^2-3\left(x^2-2x\right)+2=0\)

Answer 1

Đặt \(x^2-2x=t\) pt trở thành

\(t^2-3t+2=0\)\(\Leftrightarrow\left(t-2\right)\left(t-1\right)=0\)

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

\(\left[\begin{matrix}\Delta_{\left(1\right)}=\left(-2\right)^2-\left(-4\left(1\cdot2\right)\right)=12\\\Delta_{\left(2\right)}=\left(-2\right)^2-\left(-4\left(1\cdot1\right)\right)=8\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x_{1,2}=\frac{2\pm\sqrt{12}}{2}\\x_{3,4}=\frac{2\pm\sqrt{8}}{2}\end{matrix}\right.\)