\(A=\left|x-2016\right|+\left|x-2017\right|\)
\(\Leftrightarrow A=\left|x-2016\right|+\left|2017-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-2016\right|+\left|2017-x\right|\ge\left|x-2016+2017-x\right|=1\)
Dấu "=" khi \(\left\{{}\begin{matrix}x-2016\ge0\\2017-x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge2016\\x\le2017\end{matrix}\right.\) \(\Rightarrow2016\le x\le2017\)
Vậy GTNN của \(A=1\) khi \(2016\le x\le2017\)
Ta có: \(\left\{{}\begin{matrix}\left|x-2016\right|\ge x-2016\\\left|x-2017\right|=\left|2017-x\right|\ge2017-x\end{matrix}\right.\)
\(\Rightarrow A=\left|x-2016\right|+\left|x-2017\right|\ge x-2016+2017-x\)
\(\Rightarrow A\ge1\)
Đẳng thức xảy ra khi và chỉ khi:\(\left\{{}\begin{matrix}\left|x-2016\right|=x-2016\\\left|2017-x\right|=2017-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2016\ge0\\2017-x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2016\\x\le2017\end{matrix}\right.\)
\(\Leftrightarrow2016\le x\le2017\)
Vậy Min A = 1 \(\Leftrightarrow2016\le x\le2017\)
