a, Ta có (x+4)(\(x^2\)+1)=0\(\Leftrightarrow\left[\begin{matrix}x+4=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-4\left(chọn\right)\\x^2=-1\left(loại\right)\end{matrix}\right.\)
Vậy x=-4
b, Ta có \(\left(\left|x\right|+2\right)\left(x^2-1\right)=0\Leftrightarrow\left[\begin{matrix}\left|x\right|+2=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}\left|x\right|=-2\left(loại\right)\\x=\pm1\left(chọn\right)\end{matrix}\right.\)