\(A=\left|x-2012\right|+\left|x-2013\right|\\ A=\left|x-2012\right|+\left|2013-x\right|\)
Có: \(\left|x-2012\right|\ge x-2012\text{ với mọi }x\)
\(\left|2013-x\right|\ge2013-x\text{ với mọi }x\)
\(\Rightarrow\left|x-2012\right|+\left|2013-x\right|\ge x-2012+2013-x\text{ với mọi }x\\ \Rightarrow A\ge1\text{ với mọi }x\)
Vậy GTNN của A = 1
\("="\Leftrightarrow\left\{{}\begin{matrix}\left|x-2012\right|=x-2012\\\left|2013-x\right|=2013-x\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-2012\ge0\\2013-x\ge0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ge2012\\x\le2013\end{matrix}\right.\\ \Leftrightarrow2012\le x\le2013\)
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\(A=\left|x-1\right|+\left|x+2012\right|\\ A=\left|1-x\right|+\left|x+2012\right|\)
Có: \(\left|1-x\right|\ge1-x\text{ với mọi }x\)
\(\left|x+2012\right|\ge x+2012\text{ với mọi }x\)
\(\Rightarrow\left|1-x\right|+\left|x+2012\right|\ge1-x+x+2012\text{ với mọi }x\\ \Rightarrow A\ge2013\text{ với mọi }x\)
Vậy GTNN của A = 2013
\("="\Leftrightarrow\left\{{}\begin{matrix}\left|1-x\right|=1-x\\\left|x+2012\right|=x+2012\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x+2012\ge0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\le1\\x\ge-2012\end{matrix}\right.\\ \Leftrightarrow-2012\le x\le1\)
