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
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）]
=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
观察下列各式,1\/1*2=1-1\/2 1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4
化简成1\/(x-1)-1\/x+1\/(x-2)-1\/(x-1)+…+1\/(x-10)-1\/(x-9)=5\/12 1\/(x-10)-1\/x=5\/12 10\/(x(x-10))=5\/12 x=12
...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...
左边=1\/1*2+1\/2*3+1\/3*4 右边=1\/1 - 1\/2 + 1\/2 - 1\/3 + 1\/3 - 1\/4 = 1- 1\/4 = 3\/4 因此：1）:1\/1*2+1\/2*3+1\/3*4+...+1\/n(n+1)=1 - 1\/(n+1)2)1\/n(n+2)= = (1\/2)* [(n+2)-n]\/n(n+2) = (1\/2)* [1\/n - 1\/(n+2)]3)1\/...
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)=...
用简便方法如何计算?1\/1×2 +1\/2×3+1\/3×4+1\/4×5
裂项求解 1\/1*2=1-1\/2 1\/2*3=1\/2-1\/3 ……1\/4*5=1\/4-1\/5 以上各式相加，结果是4\/5
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，...
请教一道小学数学题,1\/1x2+1\/2x3+1\/3x4+1\/4x5+...1\/2005x2006怎么算...
解:根据两个连续自然数A,B.1\/A-1\/B=1\/AB得 原式 =1-1\/2+(1\/2-1\/3)+(1\/4-1\/3)+(1\/5-1\/4)+...+(1\/2006-1\/2005)=1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/5+...+1\/2005-1\/2006)=1-1\/2006 =2005\/2006....
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！
