\(A=\left|x-1\right|+\left|x+2012\right|=\left|x-1\right|+\left|2012-x\right|\)
Với mọi x ta có :
\(A=\left|x-1\right|+\left|2012-x\right|\)
\(\Leftrightarrow A\ge\left|x-1+2012-x\right|\)
\(\Leftrightarrow A\ge\left|2011\right|\)
\(\Leftrightarrow A\ge2011\)
Dấu "=" xảy ra khi :
\(\left(x-1\right)\left(2012-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\2012-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\2012-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\2012\le x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le1\\2012\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}1\le x\le2012\\x\in\varnothing\end{matrix}\right.\)
Vậy ..
Ta có: \(\left\{{}\begin{matrix}\left|x-1\right|=\left|1-x\right|\ge1-x\\\left|x+2012\right|\ge x+2012\end{matrix}\right.\)
\(\Rightarrow\left|x-1\right|+\left|x+2012\right|\ge\left(1-x\right)+\left(x+2012\right)\)
\(\Rightarrow A\ge2013\)
Dấu "=" xảy ra khi và chỉ khi\(\left\{{}\begin{matrix}\left|1-x\right|=1-x\\\left|x+2012\right|=x+2012\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x+2012\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x\ge-2012\end{matrix}\right.\)
\(\Leftrightarrow-2012\le x\le1\)
Vậy Min A = 2013 \(\Leftrightarrow-2012\le x\le1\)
