化简1/1*2=1-1/2,1/2*3=1/2-1/3,1/3*4=1/3-1/4则1/(x-1)x+1/x(x+1)+1/(x+1)(x+2)+..+1/(x+9)(x+10)=

如题所述

第1个回答  2019-03-03
因为不能确定a与b之间的关系
如果a=b时
1/ab
1/(a
1)(b
1)
1/(a
2)(b
2)
......1/(a
2005)(b
2005)
=1/a²
1/(a
1)²
1/(a
2)²
...
1/(a
2005)²
如果a≠b时
1/ab
1/(a
1)(b
1)
1/(a
2)(b
2)
......1/(a
2005)(b
2005)
=1/(b-a)[1/a-1/b
1/(a
1)-1/(b
1)
1/(a
2)-1/(b
2)
...
1/(a
2005)-1/(b
2005)]
第2个回答  2020-02-22
原式=1/(x-1)-1/x+1/x-1/(x-2)+1/(x-2)-,,,,,,+1/(x+9)-1/(x+10)

=1/(x-1)-1/(x+10)
分解开来后除了首尾两项之外全都消去了,可得结果

化简1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4则1\/(x-1)x+1\/x(x+1...
原式=1\/(x-1)-1\/x+1\/x-1\/(x-2)+1\/(x-2)-,,,+1\/(x+9)-1\/(x+10)=1\/(x-1)-1\/(x+10)分解开来后除了首尾两项之外全都消去了,可得结果

...等式:1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4,将以上三个等式两边...
1\/1*2 = (2-1)\/1*2 = 2\/1*2 - 1\/1*2 = 1\/1 - 1\/2 1\/2*3 = (3-2)\/2*3 = 3\/2*3 - 2\/2*3 = 1\/2 - 1\/3 1\/3*4 = (4-3)\/3*4 = 4\/3*4 - 3\/3*4 = 1\/3 - 1\/4 上述各式相加:左边=1\/1*2+1\/2*3+1\/3*4 右边=1\/1 - 1\/2 + 1\/2 - ...

已知:1\/1×2=1-1\/2;1\/2×3=1\/2-1\/3;1\/3×4=1\/3-1\/4;...
答:1\/(1×3)+1\/(3×5)+1\/(5×7)+...+1\/[(2n-1)×(2n+1)]=17\/35 两边同乘以2 2\/(1×3)+2\/(3×5)+2\/(5×7)+...+2\/[(2n-1)×(2n+1)]=34\/35 1-1\/3+1\/3-1\/5+1\/5-1\/7+...+1\/(2n-1)-1\/(2n+1)=34\/35 1-1\/(2n+1)=34\/35 1\/(2n+1)=...

1\/1*2+1\/2*3+1\/3*4+...+1\/n(n+1)化简
1、可以分析数列的规律:1\/1×2=1-1\/2,1\/2×3=1\/2-1\/3;即每个数字都可以进行拆分为两个分数相减,通项公式为:1\/n(n+1)=1\/n-1\/n+1 2、1\/1×2+1\/2×3+1\/3×4+...1\/n(n+1)=1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/n-1\/n+1=1-1\/n+1=n\/n+1。找规律填空的意义,...

1\/2=1\/1×2=1\/1-1\/2 1\/6=1\/2×3=1\/2-1\/3 1\/12=1\/3×4=1\/3-1\/4...
1\/2=1\/1×2=1\/1-1\/2 1\/6=1\/2×3=1\/2-1\/3 1\/12=1\/3×4=1\/3-1\/4 ···1.求规律1\/n(n+1)=1\/n-1\/(n+1)2.利用规律计算 1\/2+1\/6+1\/12···+1\/(n-1)n +1\/(n+1)n =1-1\/2+1\/2-1\/3+1\/3-1\/4+……+1\/(n-1)-1\/n+1\/n-1\/(n+1)=1-1\/...

1 \/1*2=1\/1-1\/2,,12*3=1\/2-1\/3,1\/3*4=1\/3-1\/4.…,1\/n*...
2009-1993=16,所以所问的结果就是1\/16×(1\/1993-1\/2009)即相差多少就除以多少。1\/[n×(n+y)]=1\/y×[1\/n-1\/(n+y)]

1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4,则1\/2 007?*2 008=...
你好:分母为相邻的两个自然数的乘积,分子为这两个自然数的差(即是1)1\/2 007*2 008=1\/2007-1\/2008 有n的式子表示你发生的规律:1\/ n*(n+1)=1\/n-1\/(n+1)祝你学习进步!

求数列1\/1x2,1\/2x3,1\/3x4,1\/4x5...的前n项和---
第n项为1\/n(n+1)由于1\/1x2=1-1\/2 1\/2x3=(1\/2)-(1\/3)1\/3x4=(1\/3)-(1\/4)……1\/n(n+1)=(1\/n)-(1\/n+1)所以前n项的和为1-(1\/n+1)

已知1-1\/2=1\/2,1\/2-1\/3=1\/6,1\/3-1\/4=1\/12,...根据这些等式解答下列各...
(1) 1\/1*2+1\/2*3+1\/3*4+1\/4*5+1\/5*6 =(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+(1\/4-1\/5)+(1\/5-1\/6)=1-1\/6 =5\/6 (2)1\/1*2+1\/2*3+1\/3*4+...+1\/n(n+1)=(1-1\/2)+(1\/2-1\/3)+……+[1\/n-1\/(n+1)=1-1\/(n+1)=n\/(n+1)(3)因为1\/1...

1\/1×2+1\/2×3+1\/3×4+...1\/99×100怎么算
1\/1×2+1\/2×3+1\/3×4+...1\/99×100 =(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/99-1\/100)把每一项都拆开来,前后抵消,最后只剩下1-1\/100=99\/100

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