Với mọi x ta có :
\(A=\left|x-2918\right|-\left|x-2017\right|\)
\(\Leftrightarrow A\ge\left|x-2018-x+2017\right|\)
\(\Leftrightarrow A\ge\left|-1\right|\)
\(\Leftrightarrow A\ge1\)
Dấu "=" xảy ra
\(\Leftrightarrow\left|x-2018\right|\ge\left|x-2017\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2018\ge0\\x-2017\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2018< 0\\x-2017< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2018\\x\ge2017\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2018\\x< 2017\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2018\\x< 2017\end{matrix}\right.\)
Vậy....
Làm CTV bên olm chán rồi,qua hoc24 kiếm GP tí!Bạn Nguyễn Thanh Hằng lạc đề r,đề là tìm gtln (do bạn đánh ngược dấu)
Ta có:
\(A\le\left|x-2018-x-2017\right|=\left|-1\right|=1\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-2018\right)\left(x-2017\right)\ge0\)
\(TH1:\left[{}\begin{matrix}x-2018\le0\\x-2017\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le2018\\x\le2017\end{matrix}\right.\Leftrightarrow x\le2017\)
TH2: \(\left[{}\begin{matrix}x-2018\ge0\\x-2017\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge2018\\x\ge2017\end{matrix}\right.\Leftrightarrow x\ge2018\)
Thử lại,dễ dàng loại TH2,vậy x =< 2017
