1/(1x2)+1/(2x3)+1/(3x4)+．．．．．．+1/(99x100)=?要计算过程．
1/(1x2)+1/(2x3)+1/(3x4)+．．．．．．+1/(99x100)=?要计算过程．
=1/1-1/2+1/2-1/3+1/3+1/4+．．．．．．+1/99-1/100
=1-1/100
=99/100本回答被提问者采纳
=1/1-1/100
=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
用简便方法计算1\/(1*2)+1\/(2*3)+…+1\/(99*100)(有过程)
1\/（1*2）+1\/（2*3）+…+1\/（99*100)=[1-1\/2]+[1\/2-1\/3]+...+[1\/98-1\/99]+[1\/99-1\/100]=1-1\/100 =99\/100
用简便方法计算1\/(1*2)+1\/(2*3)+…+1\/(99*100)(有过程)
1\/（1*2）+1\/（2*3）+…+1\/（99*100)=[1-1\/2]+[1\/2-1\/3]+...+[1\/98-1\/99]+[1\/99-1\/100]=1-1\/100 =99\/100
1\/1X2+1\/2X3+1\/3X4+...+1\/99X100 怎么简便计算。。过程..
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...
小学奥数题1\/(1*2)+1\/(2*3)+1\/(3*4)+...+1\/(99*100)
1\/(1*2)+1\/(2*3)+1\/(3*4)+1\/(4*5)+1\/(5*6)+……+1\/(98*99)+1\/(99*100)=1-1\/2+1\/2-1\/3+...+1\/99-1\/100 =1-1\/100 =99\/100
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
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
数学计算题:1\/1X2+1\/2X3+1\/3X4+...+1\/99x100=? 请写出计算过程~
1\/1*2=1\/1-1\/2 1\/2*3=1\/2-1\/3……所以原式=1-1\/2+1\/2-1\/3+1\/3-1\/4+……-1\/99+1\/99-1\/100=99\/100
一道数学题急 1\/1x2+1\/2x3+1\/3x4+...+1\/99x100=?
1\/1x2+1\/2x3+1\/3x4+...+1\/99x100 =1-1\/2+1\/2-1\/3+.+1\/98-1\/99+1\/99-1\/100 =1-1\/100 =99\/100
1\/(1×2)+1\/(2×3)+...1\/(99×100)的值
您好！因为1\/2=1-1\/2 1\/6=1\/2-1\/3 1\/12=1\/3-1\/4 依此类推，原式就＝ 1－1\/2+1\/2-1\/3+1\/3-1\/4...+1\/99-1\/100= 1-1\/100=99\/100
