Ta có :
\(A=\left|x-2012\right|+\left|x-2013\right|=\left|x-2012\right|+\left|2013-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có :
\(A=\left|x-2012\right|+\left|2013-x\right|\)
\(\Leftrightarrow A\ge\left|x-2012+2013-x\right|\)
\(\Leftrightarrow A\ge1\)
Dấu"=" xảy ra khi :
\(\left(x-2012\right)\left(2013-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2012\ge0\\2013-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2012\le0\\2013-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2012\\2013\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2012\\2013\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2012\le x\le2013\\x\in\varnothing\end{matrix}\right.\)
Vậy ...
Ta thấy:\(|\)x-2012\(|\)\(\ge\)0 vs mọi x\(\in\)Z
\(|\)y-2013\(|\)\(\ge\)0 vs mọi x\(\in\)Z
\(|\)x-2012\(|\)+\(|\)y-2013\(|\)\(\ge\)0 vs mọi x\(\in\)Z
Vậy GTNN của A\(\ge\)0
Dấu " = " xảy ra khi :
\(|\)x-2012\(|\) = 0
x-2012 = 0
x = 2012
\(|\)y-2013\(|\) = 0
y-2013 = 0
y = 2013
Vậy x = 2012; y = 2013
