已知/ab-2/+/b+1/的平方=0,求1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)+...+1/(a-2008)(b-2008)的值
问题描述：
已知/ab-2/+/b+1/的平方=0,求1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)+...+1/(a-2008)(b-2008)的值
答
因为/ab-2/+/b+1/的平方=0 所以 ab=2,b=-1 即a=-2,b=-1原式=1/2+1/(3×2)+1/(4×3)+...+1/(3000×2009) =1/2+1/2-1/3+1/3-1/4+...+1/2008-1/2009+1/2009-1/3000 =1-1/3000 ...