1/1×2 = 1 - 1/2
1/2×3 = 1/2 - 1/3
依此类推
1/99×100 = 1/99 - 1/100
所以
原式 = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/99 - 1/00
= 1 - 1/100
= 99/100
一定要帮我啊!(1\/1x2)+(1\/2x3)+(1\/3x4)+(1\/4x5)+.+(1\/99x100)=?
所以 原式 = 1 - 1\/2 + 1\/2 - 1\/3 + 1\/3 - 1\/4 + ...+ 1\/99 - 1\/00 = 1 - 1\/100 = 99\/100
1\/(1x2)+1\/(2x3)+1\/(3x4)+...+1\/(99x100)=?要计算过程.
1\/(1x2)+1\/(2x3)+1\/(3x4)+．．．+1\/(99x100)=1\/1-1\/2+1\/2-1\/3+1\/3+1\/4+．．．+1\/99-1\/100 =1-1\/100 =99\/100
1\/1X2+1\/2X3+1\/3X4+…… +1\/99X100=?
1\/1X2+1\/2X3+1\/3X4+...+1\/99X100 =（1-1\/2）+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/99-1\/100)=1-1\/100 =99\/100 乘法分配律 简便计算中最常用的方法是乘法分配律。乘法分配律指的是ax(b+c)=axb+axc其中a,b,c是任意实数。相反的，axb+axc=ax(b+c)叫做乘法分配律的逆运用...
1\/1*2+1\/2*3+1\/3*4+.+1\/99*100怎样简便运算?
=1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/99-1\/100 =1-1\/100 =99\/100 =100分之99
1\/1X2+1\/2x3+1\/3x4+1\/4x5+...+1\/98x99+1\/99x100求结果!!!
1\/1乘2等于1\/1减1\/2 同理2乘3分之一等于2分之一减去3分之一 上式等于1\/1-1\/2+1\/2--1\/3+1\/3...-1\/99+1\/99-1\/100等于99\/100
1\/2X1\/3+1\/3X1\/4+1\/4X1\/5+1\/5X1\/6+...+1\/99X1\/100=? 求解题过程,谢谢...
= 1\/(2 X 3) + 1\/(3 X 4) + 1\/(4 X 5) + ...+ 1\/(99 X 100)= (3 - 2)\/(2 X 3) + (4 - 3)\/(3 X 4) + (5 - 4)\/(4 X 5) + ...+ (100 - 99)\/(99 X 100)= (1\/2 - 1\/3) + (1\/3 - 1\/4) + (1\/4 - 1\/5) +...+ (1\/99 - 1\/...
1\/1X2+1\/2X3+1\/3X4+1\/4X5+...+1\/1999X2000=? 请告知规律.
1\/1×2+1\/2×3+1\/3×4+1\/4×5+...+1\/1999×2000=1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+…+1\/1999-1\/2000=1-1\/2000=1999\/2000其规律是：1\/1×2=1\/2,而1-1\/2=1\/2,即1\/1×2=1-1\/2；1\/2×3=1\/6,而1\/2-1\/3=1\/6,即1\/2×3=1\/2-1\/3...
数奥题1\/(1x2)+1\/(2x3)+1\/(3x4)+...+1\/(99x100)答案
这道题的分子都是1 通过观察可以发现 1\/2=1-1\/2 1\/6=1\/2-1\/3 1\/12=1\/3-1\/4 原式＝1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4+……+1\/98-1\/100 ＝99\/100 这道题的解法叫做裂项或者拆分 是初一学习的内容
简便计算1\/2+1\/2x3+1\/3x4+1\/4x5+...+1\/99x100
1\/2+1\/2x3+1\/3x4+1\/4x5+...+1\/99x100 = 1\/2 + (1\/2 - 1\/3)+(1\/3 - 1\/4) +... +(1\/99 - 1\/100)= 1\/2 +1\/2 -1\/3+1\/3 -1\/4+1\/4+...+1\/99 - 1\/100 = 1 - 1\/100 =99\/100 (*^__^* *^__^* *^__^*)有什么不明白可以对该题继续追问，随...
数奥题1\/(1x2)+1\/(2x3)+1\/(3x4)+...+1\/(99x100)答案
=1-1\/2+1\/2-1\/3+1\/3+...-1\/100[ps:1\/n*(n+1)=1\/n-1\/(n+1)]=1-1\/100 =99\/100
