\(B=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|x-5\right|\)
\(=\left|x-1\right|+\left|x-2\right|+\left|3-x\right|+\left|5-x\right|\)
\(\ge x-1+x-2+3-x+5-x=5\)
Dấu "=" khi \(\begin{cases}x-1\ge0\\x-2\ge0\\3-x\ge0\\5-x\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge1\\x\ge2\\x\le3\\x\le5\end{cases}\)\(\Leftrightarrow2\le x\le3\)
Vậy với \(2\le x\le3\) thì B đạt GTNN là 5