Ta có : \(P\left(x\right)=\left|x-2015\right|+\left|x-2016\right|+\left|x-2017\right|\)
\(=\left|x-2016\right|+\left[\left|x-2015\right|+\left|2017-x\right|\right]\)
Vì : + ) \(\left|x-2016\right|\ge0\)
+ )\(\left|x-2015\right|+\left|2017-x\right|\ge\left|x-2015+2017-x\right|=2\)
Vậy \(Min_P=2\)
Dấu "=" xảy ra khi : \(\left\{{}\begin{matrix}x-2015\ge0\\x-2016=0\\x-2017\le0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ge2015\\x=2016\\x\le2017\end{matrix}\right.\Rightarrow x=2016\)
P(x) =|x−2015|+|x−2016|+|x−2017|
=(|x−2015|+|x−2017|)+|x−2016|
Ta có: |x−2015|+|2017-x|\(\ge\)|x-2015+2017-x|=2
Dấu "=" xảy ra khi:
(x−2015).(2017-x)\(\ge\)0
\(\left\{{}\begin{matrix}x-2015\le0\\2017-x\le0\end{matrix}\right.=>\left\{{}\begin{matrix}x\le2015\\x\ge2017\end{matrix}\right.\)
=>\(2017\le x\le2015\)(VL)
\(\left\{{}\begin{matrix}x-2015\ge0\\2017-x\ge0\end{matrix}\right.=>\left\{{}\begin{matrix}x\ge2015\\x\le2017\end{matrix}\right.\)
=>\(2017\ge x\ge2015\)(TM) (1)
Mặt khác: |x−2016|\(\ge\)0
Dấu "=" xảy ra khi: x-2016=0
<=>x=2016 (2)
Từ (1) và (2) ,ta có:
P(x) \(\ge2+0=2\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}2017\ge x\ge2015\\x=2016\end{matrix}\right.=>x=2016}\)
vậy min P(x)=2 khi x=2016
