\(\left|x+3\right|+\left|x-2\right|+\left|x-5\right|\)
\(\ge x+3+0+5-x=8\)
\(\Rightarrow P\ge8\)
Dấu = khi \(\begin{cases}x+3\ge0\\x-2=0\\x-5\le0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge-3\\x=2\\x\le5\end{cases}\)\(\Leftrightarrow x=2\)
Vậy Pmin=8 khi x=2