诺|a-1|+|ab-2|=0,求1/ab+1/(a+1)(b+10)+1/(a+2)(b+2)+......+1/(a+2009)(b+2009)的值
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a=1.ab=2.所以：b=2.
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+......+1/(a+2009)(b+2009)
=1/1*2+1/(1+1)(2+1)+1/(1+2)(2+2)+......+1/(1+2009)(2+2009)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/2010-1/2011)
=1-1/2011
=2010/2011
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+......+1/(a+2009)(b+2009)
=1/1*2+1/(1+1)(2+1)+1/(1+2)(2+2)+......+1/(1+2009)(2+2009)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/2010-1/2011)
=1-1/2011
=2010/2011
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