解：x=2又2013/2014所以x>0 x-1>0 x-2>0 x-3<0 x-4<0 x-5<0于是|x|+|x-1|+|x-2|+|x-3|+|x-4|+|x-5|=x+x-1+x-2-(x-3)-(x-4)-(x-5)=x+x-1+x-2-x+3-x+4-x+5=9
2|x|+|x-1|+|x-2|+|x-3|+|x-4|+|x-5|=x+x-1+x-2+3-x+4-x+5-x=-1-2+3+4+5=9
