∵|ab-2|≥0,(1-b) 2≥0,且|ab-2|+(1-b) 2=0,
∴ab-2=0,且1-b=0,解得ab=2,且b=1,
把b=1代入ab=2中,解得a=2,
则
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
2008 -
1
2009
=1-
1
2009
=
2008
2009 .