\(\left(x^2-x+1\right)\left(x^2+4x+1\right)=6x^2\)
Đặt \(x^2-x+1=t\left(t\ge\dfrac{3}{4}\right)\)
\(\Rightarrow t\left(t+5x\right)=6x^2\)
\(\Leftrightarrow t^2+5xt-6x^2=0\)
\(\Leftrightarrow\left(t+6x\right)\left(t-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-6x\\t=x\end{matrix}\right.\)
\(\odot\) TH1: \(t=-6x\)
\(\Rightarrow x^2-x+1=-6x\)
\(\Leftrightarrow x^2+5x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{21}}{2}\\x=\dfrac{-5-\sqrt{21}}{2}\end{matrix}\right.\)
\(\odot\) TH2: \(t=x\)
\(\Rightarrow x^2-x+1=x\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow x=1\)
Vậy phương trình đã cho có tập nghiệm \(S=\left\{1;\dfrac{-5+\sqrt{21}}{2};\dfrac{-5-\sqrt{21}}{2}\right\}\)
