1\/1x2十1\/2x3十1\/3x4十…十1\/98x99十1\/99x100=?
1\/1x2十1\/2x3十1\/3x4十…十1\/98x99十1\/99x100 =1-1\/2+1\/2-1\/3+1\/3-1\/4+……+1\/98-1\/99+1\/99-1\/100 =1-1\/100 =99\/100
1\/1x2十1\/2x3十1\/3x4十…十1\/98x99十1\/99x100=?
这是典型的裂项求和。1\/[x(x+1)]=1\/x-1\/(1+x),然后用累加法。给你提供一个结论：Sn=1\/1x2十1\/2x3十1\/3x4十…十1\/[x(x+1)]=x\/(x+1)。所以这一题的答案是99\/100.如果还有什么不懂的请追问。
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\/1x2+1\/2x3+1\/3x4+...1\/99x100
解：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\/2+1\/2-1\/3+1\/3-1\/4+...+1\/99-1\/100 （中间两项两项正负抵消）=1-1\/100 =99\/100
1\/1x2+1\/2x3+1\/3x4+...+1\/98x99+1\/99x100
原等式=（1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/98-1\/99)+(1\/99-1\/100)=99\/100
1\/1X2+1\/2X3+1\/3X4...1\/99X100
1\/1X2+1\/2X3+1\/3X4+...+1\/99X100 = 1\/1-1\/2+1\/2-1\/3+1\/3-...+1\/99-1\/00 = 1-1\/100 = 99\/100
1\/1X2+1\/2X3+1\/3X4+...+1\/99+100=
题目给错了吧？ 最后一项应该是1\/99X100吧？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
1X2\/1+2X3\/1+3X4\/1…98X99\/1+99X100\/1=?
我写’\/‘前分子分母 1\/1X2+1\/2X3+1\/3X4…1\/98X99+1\/99X100 =(1\/1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/98-1\/99)+(1\/99-1\/100)=1-1\/100 去括号间项抵消 =99\/100 公式 1\/[n*(n-1)]=1\/n - 1\/(n-1)
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
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
