Question

如果有理数a，b满足|ab-2|+(1-b)^2=0，求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2007）(b+2007)的值：

Answer 1

由已知得b=1 a=2
所以式子等于1/(1*2)+1/(2*3)+1/(3*4)+.....+1/(2008*2009）=
(1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/2008-1/2009)=2008/2009

Answer 2

由题意得（1-b)^2=0 |ab-2|=0
1-b=0 b=1
ab-2=0 ab=2
b=2
原式=1/2*1+1/3*2+1/4*3+……+1/2009*2008
=1-1/2+1/2-1/3+1/3-1/4+……-1/2009
=1-1/2009
=2008/2009