Question

[Nguyễn Thu Thủy]

tìm min A=/2x-2017/+/2x-2018/

Answer 1

[ɱ√ρ︵ʙᴇsт➻❥ɴᴀκᴀʀoтн★彡]

Ta có: 

\(\left|2x-2017\right|\ge2x-2017\forall x\)

\(\left|2x-2018\right|=\left|2018-2x\right|\ge2018-2x\forall x\)

\(\Rightarrow\left|2x-2017\right|-\left|2x-2018\right|\ge1\)

Dấu "=" xảy ra khi 

\(\hept{\begin{cases}2x-2017\ge0\\2x-2018\le0\end{cases}\Rightarrow\hept{\begin{cases}x\ge\frac{2017}{2}\\x\le\frac{2018}{2}\end{cases}}}\)