因为|ab-2|≥0
|a-1|≥0
且|ab-2|与|a-1|互为相反数
所以|ab-2|=|a-1|=0
所以a=1,b=2.
所以1/ab+1/(a+1)(b+1)+...+1/(a+2006)(b+2006)
=1/2+1/2*3+...+1/2007*2008
=1/2+1/2-1/3+1/3-1/4+...+1/2007-1/2008
=1-1/2008
=2007/2008
因为|ab-2|≥0
|a-1|≥0
且|ab-2|与|a-1|互为相反数
所以|ab-2|=|a-1|=0
所以a=1,b=2.
所以1/ab+1/(a+1)(b+1)+...+1/(a+2006)(b+2006)
=1/2+1/2*3+...+1/2007*2008
=1/2+1/2-1/3+1/3-1/4+...+1/2007-1/2008
=1-1/2008
=2007/2008