Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có:
\(A=\left|2x-2\right|+\left|2x-2013\right|=\left|2x-2\right|+\left|2013-2x\right|\ge\left|2x-2+2013-2x\right|=2011\)Vậy Min A=2011
Dấu "=" xảy ra khi \(\left(2x-2\right)\left(2013-x\right)\ge0\Rightarrow1\le x\le\frac{2013}{2}\)
Vậy Min A=2011\(\Leftrightarrow1\le x\le\frac{2013}{5}\)