Ta có :
\(A=\left|x-2010\right|+\left|x+10\right|=\left|x-2010\right|+\left|10-x\right|\ge\left|x-2010+10-x\right|=\left|-2000\right|=2000\)
Dấu "=" xảy ra khi :
\(\left(x-2010\right)\left(10-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2010\ge0\\10-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2010\le0\\10-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2010\\10\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2010\\10\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\2010\ge x\ge10\end{matrix}\right.\)
Vậy \(A_{Min}=2000\Leftrightarrow2010\ge x\ge10\)
