|ab-2|+(b-1)^2=0
所以b-1=0,ab-2=0
b=1,ab=2
a=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
如果|ab-2|+(b-1)^2=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+~+1/(a+20
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