Ta có : \(A=\left|2x-1\right|+\left|2x-2013\right|=\left|2x-1\right|+\left|2013-2x\right|\)
\(\Rightarrow A\ge\left|2x-1+2013-2x\right|=\left|2012\right|=2012\)
Dấu "=" xảy ra <=> \(\left(2x-1\right)\left(2013-2x\right)\ge0\Rightarrow\frac{1}{2}\le x\le\frac{2013}{2}\)
Vậy \(A_{min}=2012\) tại \(\frac{1}{2}\le x\le\frac{2013}{2}\)