\(A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+...+\left|x-2015\right|\)
Ta có : \(A\) có : \(\left(2015-1\right):1+1=2015\) (số hạng)
\(A=\left|x-1\right|\ge x-1\)
\(\left|x-2\right|\ge x-2\)
\(\left|x-3\right|\ge x-3\)
\(..........\)
\(\left|x-1007\right|\ge x-1007\)
\(\left|x-1008\right|\ge0\)
\(\left|x-1009\right|\ge-x+1009\)
\(\left|x-1010\right|\ge-x+1010\)
\(\left|x-1011\right|\ge-x+1011\)
\(..........\)
\(\left|x-2015\right|\ge-x+2015\)
\(\Rightarrow\)\(A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+...+\left|x-2015\right|\ge x-1+x-2+x-3+...+x-1007+0-x+1009-x+1010-x+1011+...+-x+2015=2+4+6+...+2014\)
\(=\dfrac{\left[\dfrac{\left(2014-2\right)}{2}+1\right].\left(2014+2\right)}{2}=1015056\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-1\ge0\\\left|x-1008\right|=0\\-x+2015\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x=1008\\x\le2015\end{matrix}\right.\)
Vậy \(\min\limits_A=1015056\Leftrightarrow x=1008\)
