\(\left(x^2-4\right)\left(2x+x+3\right)=0\\ =>\left[{}\begin{matrix}x^2-4=0\\3x+3=0\end{matrix}\right.=>\left[{}\begin{matrix}x^2=4\\x=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\x=-2\\x=-1\end{matrix}\right.\)
\(\left(x^2-4\right)\left(2x+x+3\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left[x\left(2+1\right)+3\right]=0\)
\(\Rightarrow\left(x^2-4\right)\left(3x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\3x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=4=\left(\pm2\right)^2\\3x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=-1\end{matrix}\right.\) Vậy \(x\in\left\{\pm2;-1\right\}\)
