若a b为实数,(绝对值a-1)+(根号ab-2)=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2008)(b+2008)的值

绝对值和根号大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立
所以两个都等于0
所以a-1=0,ab-2=0
a=1,ab=2
b=2
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2008)(b+2008)
=1/1*2+1/2*3+1/3*4+...+1/2009*2010
=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2009-1/2010)
=1/1-1/2010
=2009/2010