\(\Leftrightarrow\left(x^2+x+3\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)+3\left(x^2+x\right)-6=0\\ \Leftrightarrow\left(x^2+x-2\right)\left(x^2+x\right)+3\left(x^2+x-2\right)=0\\ \Leftrightarrow\left(x^2+x+3\right)\left(x^2+x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0\left(vô.n_0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(\left(x^2+x\right)^2+\left(x^2+x\right)-6=0\)
Đặt \(x^2+x=t\left(t\ge-\dfrac{1}{4}\right)\)
Pt\(\Rightarrow t^2+t-6=0\)\(\Rightarrow\left[{}\begin{matrix}t=2\left(tm\right)\\t=-3\left(loại\right)\end{matrix}\right.\)
Thay \(t=2\) vào \(x^2+x=t\) ta đc:
\(x^2+x=2\Rightarrow x^2+x-2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
