$A=|x-1|+|x+2012|\\=|x-1|+|-x-2012|\\\Rightarrow A\ge |x-1-x-2012|=2013$
Dấu "=" xảy ra khi $(x-1)(-x-2012)\ge 0$
$\Leftrightarrow -2012\le x\le 1$
Vậy $\min A=2013$ khi $-2012\le x\le 1$
\(A=\left|x-1\right|+\left|x+2012\right|=\left|x-1\right|+\left|-2012-x\right|\ge\left|x-1-2012-x\right|=\left|-2013\right|=2013\)
\(minA=2011\Leftrightarrow\left(x-1\right)\left(-x-2012\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\-x-2012\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\-x-2012\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\x\le-2012\end{matrix}\right.\\\left\{{}\begin{matrix}x\le1\\x\ge-2012\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow-2012\le x\le1\)
