5) \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)
Đặt \(x^2+x+1=a\) Ta được:
\(a.\left(a+1\right)=12\)
\(\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow\left(a^2+4a\right)-\left(3a+12\right)=0\)
\(\Leftrightarrow a\left(a+4\right)-3\left(a+4\right)=0\)
\(\Leftrightarrow\left(a+4\right)\left(a-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+4=0\\a-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-4\\a=3\end{matrix}\right.\)
* \(a=-4\)
\(\Leftrightarrow x^2+x+1=-4\)
\(\Leftrightarrow x^2+x+5=0\) ( vô nghiệm, loại )
* \(a=3\)
\(\Leftrightarrow x^2+x+1=3\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left(x^2-x\right)+\left(2x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{1;-2\right\}\)
