\(P=\left|x-2011\right|+\left|x-1\right|=\left|2011-x\right|+\left|x-1\right|\ge\left|2011-x+x-1\right|=2010\)
\(\Rightarrow MIN_P=2010\Leftrightarrow\left(2011-x\right)\left(x-1\right)\ge0\)
\(\Leftrightarrow1\le x\le2011\)
Vậy MINP=2010 khi \(1\le x\le2011\)