\(x\left(x+2\right)\left(x^2+2x+5\right)=6\)
\(\Leftrightarrow\left(x^2+2x\right)\left(x^2+2x+5\right)-6=0\)
Đặt \(t=x^2+2x\) thì ta có:
\(t\left(t+5\right)-6=0\)
\(\Leftrightarrow\left(t-1\right)\left(t+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t-1=0\\t+6=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+2x=1\\x^2+2x=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x^2+2x-1=0\\x^2+2x+6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+2x+1-2=0\\x^2+2x+1+5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2-2=0\\\left(x+1\right)^2+5>0\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2=2\Rightarrow x+1=\pm\sqrt{2}\Rightarrow x=\pm\sqrt{2}-1\)