a,b满足根号a-1+根号b-2=0 求1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+.+1/(a+20？
1个回答
展开全部
√a-1+√b-2=0
a=1,b=2
1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)
=1/2+1/2*3+1/3*4+1/4*5+...+1/2009*2010
=1-1/2+1/2-1/3+1/3-1/4+...+1/2009-1/2010
=1-1/2010
=2009/2010,4,a,b满足根号a-1+根号b-2=0 求1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)的值 要用的
a,b满足根号a-1+根号b-2=0 求1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)的值
a=1,b=2
1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)
=1/2+1/2*3+1/3*4+1/4*5+...+1/2009*2010
=1-1/2+1/2-1/3+1/3-1/4+...+1/2009-1/2010
=1-1/2010
=2009/2010,4,a,b满足根号a-1+根号b-2=0 求1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)的值 要用的
a,b满足根号a-1+根号b-2=0 求1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)的值
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
