Question

已知|ab-2|+|a-2|=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.....+1/(a+2009)(b+2009) 求值？

急

Answer 1

解：由|ab-2|+|a-2|=0 可知 ab=2 a-2=0 ‘’a=2 ‘’b=1
绝对值为非负数，和为0只有分别等于零
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.....+1/(a+2009)(b+2009)
=1/2*3+1/3*2+1/4*3+.....+1/2011*2010
=1-1\2 + (1\2 -1\3) + (1\3-1\4 )+.....+(1\2010-1\2011)
=1-1\2011
=2010\2011
不容易啊 ！给分吧 如果还有什么需要问的加我好友 9902799 QQ上给你说，是在考试吧

追问

把那个电子档传我

Answer 2

绝对值都是非负数,两个非负数的和为0,那么这两个数都是0
ab-2=0
a-2=0
解得:a=2,b=1

1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.....+1/(a+2009)(b+2009)
=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(2010*2011)
=1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011
=1-1/2011
=2010/2011

Answer 3

因为√a-1+（ab-2）^2=0,
所以√a-1=0,(ab-2）^2=0,
所以a-1=0,ab-2=0
所以a=1,b=2
则1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.....+1/(a+2009)(b+2009)
=1/(1*2)+1/(2*3)+1/(3*4)+..+1/(2010*2011)
=1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011
=1-1/2011
=2010/2011

Answer 4

绝对值为非负数，和为0只有分别等于零
ab-2=0,a-2=0
a=2
2b-2=2
b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.....+1/(a+2009)(b+2009) 1/1×2+1/2×3+...+1/2010×2011
=1-1/2+1/2-1/3+1/3-1/4+...+1/2009-1/2010+1/2010-1/2011
=1-1/2011
=2010/2011

Answer 5

已知Iab-2I+Ia-2I=0
满足上式的条件是
ab-2=0
a-2=0
解得 a=2 b=1
所以原式=1/1*2+1/2*3+1/3*4+...+1/2010*2011
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2010-1/2011)
=1-1/2011 (中间的都减掉)
=2010/2011
希望能帮到你O(∩_∩)O

Answer 6

|ab-2|+|a-2|=0
根据绝对值大于等于0，两个绝对值的和等于0，那么每个绝对值都等于0
所以：ab-2=0 a-2=0
a=2 , b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.....+1/(a+2009)(b+2009)
=1/(1*2)+1/(2*3)+1/(3*4)+......+1/(2010*2011)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/2009-1/2010)+(1/2010-1/2011)
=1-1/2+1/2-1/3+1/3-1/4+......+1/2009-1/2010+1/2010-1/2011

楼主应该看到这里的规律：就是分母是两个相连的数字的乘积。根据1/[a*(a+1)]=1/a-1/(a+1)
所以得到了上式的化简。然后看到从第二项开始，依次可以抵消，抵消之后最后只剩下
1-1/2011
所以上式=1-1/2011=2010/2011
希望帮助到你，望采纳，谢谢！