已知（a-1)的平方+ab-2的绝对值=0,求1/ab+1/(a+1)*(b+1)+1/(a+2)*(b+2)+…+1/

由题意a-1=0,ab-2=0
则a=1,b=2
则原式=1/2+1/2*(1/3)+1/3*(1/4)+……+1/2011*(1/2012)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2011-1/2012)
=1-1/2012
=2011/2012

因已知（a-1)的平方+ab-2的绝对值=0，所以a-1=0,a=1;ab-2的绝对值=0,ab=2,b=2
1/ab+1/(a+1)*(b+1)+1/(a+2)*(b+2)+…+1/(a+2009)*(b+2009)+1/(a+2010)*(b+2010)
=1/2+1/(2*3)+1/(3*4)+......+1/(2010*2011)+1/(2011*2012)
=1/2+1/2-1/3+1/3-1/4+......+1/2010-1/2011+1/2011-1/2012
=1-1/2012
=2011/2012