Question

已知(a-1)2次方+ab-2的绝对值=0

1/ab加
1/(a+1)(b+1)+1/(a+2)(b+2)+......+1/(a+2009)(b+2009)

Answer 1

（a-1）^2+|ab-2|=0
则a-1=0
ab-2=0
即a=1 b=2
所以
1/ab+1/(a+1）（b+1）+1/(a+2)(b+2)+……+1/（a+2009）(b+2009)
=1/1x2+1/2x3+1/3x4+……+1/2010x2011
=1-1/2+1/2-1/3+1/3-1/4+……+1/2010-1/2011
=1-1/2011
=2010/2011

Answer 2

(a-1)2次方+ab-2的绝对值=0

(a-1)2次方=0，ab-2的绝对值=0

a-1=0,ab-2=0
a=1,b=2
所以
1/ab加
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+....+1/2010-1/2011
=1-1/2011
=2010/2011

追问

可不可以简单一点

追答

可以
(a-1)2次方+ab-2的绝对值=0

所以
a=1,b=2
1/ab加
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+....+1/2010-1/2011
=1-1/2011
=2010/2011