\(\left(x^2-3x\right)^2-5\left(x^2-3x\right)+4=0\)
Đặt \(x^2-3x=a\) ta được:
\(a^2-5a+4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-3x=1\\x^2-3x=4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-3x-1=0\\x^2-3x-4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{3+\sqrt{13}}{2}\\x=\frac{3-\sqrt{13}}{2}\\x=-1\\x=4\end{matrix}\right.\)
Đúng 0
Bình luận (0)
