\(\left|4-x^2\right|+\left|2x+2x^2\right|=4\)
\(\left|x^2-4\right|+\left|2x^2+2x\right|=4\)
\(\left\{\begin{matrix}x\le-2\\x^2-4+2x^2+2x=4\Rightarrow\left(x+2\right)\left(3x-4\right)\Rightarrow\left[\begin{matrix}x=-2\\x=\dfrac{4}{3}\left(loai\right)\end{matrix}\right.\end{matrix}\right.\)
\(\left\{\begin{matrix}-2< x\le-1\\4-x^2+2x^2+2x=4\Rightarrow x^2+2x=0\Rightarrow\left[\begin{matrix}x=0\\x=-2\end{matrix}\right.\end{matrix}\right.\)(loại)
\(\left\{\begin{matrix}-1< x< 0\\4-x^2-2x^2-2x=4\Rightarrow\left\{\begin{matrix}x=0\\x=-\dfrac{2}{3}\left(nhan\right)\end{matrix}\right.\end{matrix}\right.\)
\(\left\{\begin{matrix}0\le x< 2\\4-x^2+2x^2+2x=4\Rightarrow\left\{\begin{matrix}x=0\left(nhan\right)\\x=-2\left(loai\right)\end{matrix}\right.\end{matrix}\right.\)
\(\left\{\begin{matrix}x\ge2\\x^2-4+2x^2+2x=4\Rightarrow\left\{\begin{matrix}x=-2\\x=\dfrac{4}{3}\end{matrix}\right.\end{matrix}\right.\)(loại)
Kết luận: \(\left\{\begin{matrix}x=-2\\x=-\dfrac{2}{3}\\x=0\end{matrix}\right.\)