已知（a-1)的绝对值+（ab-2)的二次方=0,求1/+1)(b+1)+1/(a+2)(b+2)+.1/(a+2008)(b+2008)的值
问题描述：
已知（a-1)的绝对值+（ab-2)的二次方=0,求1/+1)(b+1)+1/(a+2)(b+2)+.1/(a+2008)(b+2008)的值
答
│a-1│+(ab-2)²=0因为绝对值和平方不能为负数,所以：a-1=0,ab-2=0.解得：a=1 b=21/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2008)(b+2008) =1/(2x3)+1/(3x4)+.1/(2009x2010) 把a=1 b=2代入 =(1/2-1/3)+(1/3-1/4)+.+(1/2...