因为(a-1)^2>=0
|ab-2|>=0
欲使(a-1)^2+|ab-2|=0
所腊辩宏以必有轮册
(a-1)^2=0
|ab-2|=0
可得:a=1,b=2
所以1/ab+1/[(a+1)(b+1)]+1/[(a+2)(b+2)]+...+1/[(a+2009)(b+2009)]
=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(2010*2011)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2010-1/2011)
=1-1/灶返2011
=2010/2011