\(\left|x+1\right|+\left|x+5\right|=4\\ Áp\text{ }dụng\text{ }bất\text{ }đẳng\text{ }thức\text{ }\left|a\right|+\left|b\right|\ge\left|a+b\right|,\text{ }ta\text{ }được:\\ \left|x+1\right|+\left|x+5\right|\ge\left|x+1+x+5\right|\\ \Leftrightarrow\left|2x+6\right|\le4\\ \Leftrightarrow\left[{}\begin{matrix}2x+6\le-4\\2x+6\le4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x\le-10\\2x\le-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le-5\\x\le-1\end{matrix}\right.\\\text{ }Vậy\text{ }x\le-5\text{ }hoặc\text{ }x\le-1\)