已知绝对值a-1+绝对值ab-2=0,求AB分之一+(A+1)(B+1)分之一+(A+2) (B+2)分之一+……+(A+2000)(B+2000)分之

算出答案是多少

|a-1| ≥ 0, |ab-2| ≥ 0

|a-1| + |ab-2| = 0

因此 a-1 = 0, ab - 2 = 0解得 a = 1, b = 2。原式化为：

1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(2001x2002)

=1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/1999 - 1/2000 + 1/2000 - 1/2001

=1 - 1/2001

=2000/2001