已知（a-1)的平方+b-2的绝对值=0,求1/ab+1/(a+1)*(b+2）...+1/(a+1998)*(b+1998)的值

(a-1)²+|b-2|=0
因为(a-1)²≥0,|b-2|≥0
要使它们的和等于0,则它们都必须等于0
所以a-1=0,b-2=0
a=1,b=2
1/ab+1/(a+1)*(b+2）...+1/(a+1998)*(b+1998)
=1/1×2+1/2×3+1/3×4+...+1/1999×2000
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/1999-1/2000)
=1-1/2+1/2-1/3+1/3-1/4+...+1/9999-1/2000
=1-1/2000
=1999/2000