∵|ab-2|+(1-b)²=0
∴|ab-2|=0,(1-b)²=0
{ab-2=0
1-b=0
解得：a=2,b=1
∴(1/ab)+[1/(a+1)(b+1)]+[1/(a+2)(b+2)]+……+[1/(a+2007)(b+2007)]
=(1/2)+[1/(3×2)]+[1/(4×3)]+……+[1/(2009×2008)]
=1-(1/2)+(1/2)-(1/3)+(1/3)-(1/4)+……+(1/2008)-(1/2009)
=1+[-(1/2)+(1/2)]+[-(1/3)+(1/3)]+[-(1/4)+(1/4)]……+[-(1/2008)+(1/2008)]-(1/2009)
=1+0+0+0+……+0-(1/2009)
=1-(1/2009)
=2008/2009