|ab-2 |+(b-1)×（b-1）=0,求ab分之一+（a+1）(b+1)分之一+...+（a+2000）(b+20

已知,|ab-2|+（b-1）² = 0 ,
因为,|ab-2| ≥ 0 ,（b-1）² ≥ 0 ,
所以,|ab-2| = 0 ,（b-1）² = 0 ,
可得：ab-2 = 0 ,b-1 = 0 ,
解得：a = 2 ,b = 1 ；
1/ab+1/( a+1)(b+1)+.+1/(a+2000)(b+2000)
= 1/(1*2)+1/(2*3)+.+1/(2001*2002)
= (1-1/2)+(1/2-1/3)+.+(1/2001-1/2002)
= 1-1/2002
= 2001/2002

由已知得a=1,b=2,所以1/(a+k)(b+k)=1/(k+1)(k+2)=1/(k+1)-1/(k+2),所以原式值为2001/2002