Ta có : \(\left|x+3\right|\ge x+3\)
\(\left|x-2\right|\ge2-x\)
\(\left|x-5\right|\ge0\)
\(\Rightarrow B=\left|x+3\right|+\left|x-2\right|+\left|x-5\right|\ge x+3+2-x=5\)
\(Min_B=5\Leftrightarrow\begin{cases}x+3\ge0\rightarrow x\ge-3\\x-2\le0\rightarrow x\le2\end{cases}.\)