已知|ab-2|与|a-1|互为相反数,求1/ab+1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……1

题目：已知|ab-2|与|a-1|互为相反数,试求下式的值:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……1/(a+2010)(b+2010)
因为|ab-2|与|a-1|互为相反数
所以|ab-2|和|a-1|都是非负数
所以必有ab-2=0,a-1=0
于是a=1,b=2
原式
=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(2011*2012)
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2011-1/2012
=1-1/2012=2011/2012

丨ab-2丨+Ia-1I=0
则，ab-2=0 a-1=0
a=1 b=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2010)(b+2010)
=1/2*1+1/3*2+1/3*4+/14*5+++++++.......+1/2012*2011
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+......