\(\left(x+2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+2x+7\right)+\left(x-2\right)\left(2x+4\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2+4x+6=0\end{matrix}\right.\)
Có: \(x^2+4x+6=x^2+2.2+2x+4+2=\left(x+2\right)^2+2\)
\(\Leftrightarrow\left(x+2\right)^2\ge0\forall x\Rightarrow\left(x+2\right)^2\ge2\forall x\Rightarrow x^2+4x+6\ge2\forall x\)
\(\Rightarrow x+2=0\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)