Question

tìm giá trị nhỏ nhất

\(\left|x-2017\right|+\left|x-2016\right|+\left|x-2015\right|+3\)

Answer 1

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(VT=\left|x-2017\right|+\left|x-2016\right|+\left|x-2015\right|+3\)

\(=\left|2017-x\right|+\left|x-2016\right|+\left|x-2015\right|+3\)

\(\ge\left|2017-x+x-2015\right|+0+3=5\)

Xảy ra khi \(\left\{{}\begin{matrix}2017-x\le0\\x-2016=0\\x-2015\ge0\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\le2017\\x=2016\\x\ge2015\end{matrix}\right.\)\(\Rightarrow x=2016\)