\(\left|x+1\right|+\left|x+2\right|+...+\left|x+9\right|\\ =\left|-x-1\right|+\left|-x-2\right|+...+\left|x+5\right|+...+\left|x+9\right|\\ \text{Áp dụng }BDT\text{ }\left|a\right|\ge a:\\ \Rightarrow\left|-x-1\right|+\left|-x-2\right|+...+\left|x+5\right|+...+\left|x+9\right|\\ \ge-x-1-x-2+...+\left|x+5\right|+...+x+9\\ =\left|x+5\right|-\left(1+2+3+4\right)+\left(6+7+8+9\right)\\ \\ =\left|x+5\right|+20\ge20\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}-\left(x+1\right)\ge0\\...\\ -\left(x+4\right)\ge0\\x+5=0\\x+6\ge0\\ ...\\ x+9\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\...x\le-4\\x=-5\\x\ge-6\\ ...\\ x\ge-9\end{matrix}\right. \\ \Leftrightarrow\left\{{}\begin{matrix}x\le-4\\x=-5\\x\ge-6\end{matrix}\right.\Leftrightarrow x=-5\)
Vậy \(GTNN\text{ }của\text{ }biểu\text{ }thức\text{ }là:20\) khi \(x=-5\)