已知|ab-2|+(b-1)的平方=0求1/ab+1/(a+b)(b+1)+1/(a+2)(2+b)+````````+1/(a+2008)(b+2008)（a+2008）（b+2008）的2008改为2009
问题描述：
已知|ab-2|+(b-1)的平方=0求1/ab+1/(a+b)(b+1)+1/(a+2)(2+b)+````````+1/(a+2008)(b+2008)
（a+2008）（b+2008）的2008改为2009
答
|ab-2|+(b-1)的平方=0
ab-2=0,b=1
a=2,b=1
1/ab+1/(a+b)(b+1)+1/(a+2)(2+b)+````````+1/(a+2008)(b+2008)
=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)
=1-1/2009
=2008/2009