\(\left(x^2+x\right)\left(x^2+x+1\right)=6\)
Đặt \(t=x^2+x\) thì ta có:
\(t\left(t+1\right)=6\Leftrightarrow t^2+t-6=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+3\right)=0\)\(\Rightarrow\left[{}\begin{matrix}t-2=0\\t+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}t=2\\t=-3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x^2+x=2\\x^2+x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+2\right)=0\\x^2+x+\dfrac{1}{4}+\dfrac{11}{4}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\end{matrix}\right.\)
