Question

Giải phương trình: \(\left(x^2-3x+1\right)\left(x^2-3x-2\right)=0\)

Nhanh nhé mik cần ngay và luôn ok

Answer 1

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}=0\\\left(x-\dfrac{3}{2}\right)^2-\dfrac{17}{4}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{3}{2}\right)^2=\dfrac{5}{4}\\\left(x-\dfrac{3}{2}\right)^2=\dfrac{17}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{5}}{2}\\x=\dfrac{3-\sqrt{5}}{2}\\x=\dfrac{3+\sqrt{17}}{2}\\x=\dfrac{3-\sqrt{17}}{2}\end{matrix}\right.\)