由题意|ab-2|+||b-1|=0
所以ab-2=0且b-1=0
得b=1,a=2
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2011)
=1/(1×2)+1/(2×3)+1/(3×4)+.+1/(2012×2013)
=1-1/2+1/2-1/3+1/3-1/4+ . + 1/2011-1/2012+1/2012-1/2013
=1-1/2013=2012/2013
使用方法为裂项相消
由题意|ab-2|+||b-1|=0
所以ab-2=0且b-1=0
得b=1,a=2
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2011)
=1/(1×2)+1/(2×3)+1/(3×4)+.+1/(2012×2013)
=1-1/2+1/2-1/3+1/3-1/4+ . + 1/2011-1/2012+1/2012-1/2013
=1-1/2013=2012/2013
使用方法为裂项相消