Ta có:
\(2\left|x+1\right|-\left(x+4\right)>0\)
\(\Leftrightarrow2\left|x+1\right|>x+4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4< 0\\\left\{{}\begin{matrix}x+4\ge0\\\left[{}\begin{matrix}2\left(x+1\right)< -\left(x+4\right)\\2\left(x+1\right)>x+4\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< -4\\\left\{{}\begin{matrix}x\ge-4\\\left[{}\begin{matrix}x< -2\\x>2\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< -4\\-4\le x\le-2\\x>2\end{matrix}\right.\)
Từ trên ta suy ra \(x\in\left(-\infty;-2\right)\cup\left(2;+\infty\right)\)
Vậy .........
