已知√(a-1)+(ab-2)²=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+........+1/(a+2008)(b+2008)的值
1个回答
展开全部
√(a-1)+(ab-2)²=0
则必有 A-1=0 AB-2=0 ==>A=1 B=2
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+........+1/(a+2008)(b+2008)
=1/2+1/(2*3)+1/(3*4)+.....+1/(2009*2010)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.....+(1/2009-1/2010)
=1-1/2010=2009/2010
则必有 A-1=0 AB-2=0 ==>A=1 B=2
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+........+1/(a+2008)(b+2008)
=1/2+1/(2*3)+1/(3*4)+.....+1/(2009*2010)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.....+(1/2009-1/2010)
=1-1/2010=2009/2010
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
