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