题目

已知|ab-2|+（b+1）^2=0
求1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)...+1/(a-2009)(b-2009)的值.

提问时间：2021-02-08

答案

已知|ab-2|+（b+1）^2=0
所以ab-2=0,b+1=0
所以b=-1,a=-2
所以
1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)...+1/(a-2009)(b-2009)
=1/(-2)(-1)+1/(-3)(-2)+1/(-4)(-3)+...+1/(-2011)(-2010)
=1/1×2+1/2×3+1/3×4+...+1/2010×2011
=1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011
=1-1/2011
=2010/2011