Ta có: \(\left|x+3\right|\ge x+3\forall x\)
\(\left|x-2\right|\ge0\forall x\)
\(\left|x-5\right|=\left|5-x\right|\ge5-x\forall x\)
\(\Rightarrow P=\left|x+3\right|+\left|x-2\right|+\left|x-5\right|\ge8\forall x\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}x+3\ge0\\x-2=0\\x-5\le0\end{cases}\Rightarrow}\hept{\begin{cases}x\ge-3\\x=2\\x\le5\end{cases}\Rightarrow}x=2\)
Vậy Pmin = 8 khi x = 2