Ta có
\(\left|x+3\right|+\left|x-2\right|+\left|x-5\right|=\left|x+3\right|+\left|x-2\right|+\left|5-x\right|\)
Vì \(\begin{cases}\left|x+3\right|\ge x+3\\\left|x-2\right|\ge0\\\left|x-5\right|\ge5-x\end{cases}\) ( với mọi x )
\(\Rightarrow\left|x+3\right|+\left|x-2\right|+\left|x-5\right|\ge\left(x+3\right)+0+\left(5-x\right)\)
\(\Rightarrow\left|x+3\right|+\left|x-2\right|+\left|x-5\right|\ge8\)
Dấu " = " xảy ra khi \(\begin{cases}x+3\ge0\\x-2=0\\5-x\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge-3\\x=2\\5\ge x\end{cases}\)\(\Leftrightarrow x=2\)
Vậy MINP=8 khi x=2
Ta đã biết \(\left|A\right|\ge A\left("="\Leftrightarrow A\ge0\right)\)
\(\left|A\right|=\left|-A\right|\) và \(\left|A\right|\ge0\left("="\Leftrightarrow A=0\right)\)
Ta có:
\(A=\left|x+3\right|+\left|x-2\right|+\left|x-5\right|\ge x+3+0+5-x=8\)
Dấu = khi \(\begin{cases}x+3\ge0\\x-2=0\\5-x\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge-3\\x=2\\x\le5\end{cases}\)\(\Rightarrow x=2\)
Vậy MinA=8 khi x=2