=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)
=1/2(1-1/3+1/3-1/5+...+1/9-1/11)
=1/2(1-1/11)
=5/11本回答被提问者采纳
1/(1*3)+1/(3*5)+1/(5*7)+1/(7*9)+1/(9*11)
＝1/2（1-1/3）+1/2（1/3-1/5）+1/2（1/5-1/7）+1/2（1/7-1/9）+1/2（1/9-1/11）
＝1/2（1-1/11）
＝5/11
原式=1/2*(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)=1/2*(1/1-1/11)=5/11
=1/2*[（1-1/3）+（1/3-1/5）+（1/5-1/7）+（1/7-1/9）+（1/9-1/11）]
=1/2*（1-1/11）
=1/2*10/11
=5/11
1\/(1*3)+1\/(3*5)+1\/(5*7)+1\/(7*9)+1\/(9*11)=?需要详细的计算步骤。
=(1\/1-1\/3+1\/3-1\/5...+1\/9-1\/11)\/2 =(1-1\/11)\/2 =5\/11
1×3\/1+3×5\/1+5×7\/1+7×9\/1+9×11\/1+11×13\/1简便运算
应该是1\/(1*3)+1\/(3*5)+1\/(5*7)+1\/(7*9)+1\/(9*11)+1\/(11*13)吧1\/(1*3)=(1\/1-1\/3)\/21\/(3*5)=(1\/3-1\/5)\/2.1\/(11*13)=(1\/11-1\/13)\/2所以 原式=(1\/1-1\/3)\/2+(1\/3-1\/5)\/2+.+(1\/11-1\/13)\/2=(1\/1-1\/13)\/2=6\/13 ...
简便运算:1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11
1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11 =1\/2*（1-1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-1\/9+1\/9-1\/11）=1\/2*（1-1\/11）=1\/2*10\/11 =5\/11
1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11的简便运算
解：令 1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11＝S 则 2S＝2\/1*3+2\/3*5+2\/5*7+2\/7*9+1\/9*11 ＝(1-1\/3)+(1\/3-1\/5)+(1\/5-1\/7)+(1\/7-1\/9)+(1\/9-1\/11)＝1-1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-1\/9+1\/9-1\/11 ＝1-1\/11 ＝10\/11 即原式＝S＝10...
简便运算:1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11
1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11 =1\/2*（1-1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-1\/9+1\/9-1\/11）=1\/2*（1-1\/11）=1\/2*10\/11 =5\/11
1\/(1*3*5)+1\/(3*5*7)+1\/(5*7*9)+1\/(7*9*11)
先通分,分母为1*3*5*7*9*11,第一式分子为7*9*11,第二式分子为1*9*11,第三式分子为1*3*11,第四式分子为1*3*5
1\/1×3+1\/3×5+1\/5×7+1\/7×9+1\/9×11 = 多少?? 过程也写详细一点不要...
因为1\/1×3=1\/1-1\/3，所以以此类推，1\/(n-2)-1\/n=1\/(nx(n-2))，所以可得 1\/1×3＋1\/3×5+1\/5×7+1\/7×9+1\/9×11 =1\/1-1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-1\/9+1\/9-1\/11=1\/1-1\/11=10\/11
分数的简便运算:1\/1×3+1\/3×5+1\/5×7+1\/7×9+1\/9×11
1\/1×3＋1\/3×5＋1\/5×7＋1\/7×9＋1\/9×11 =1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-1\/9+1\/9-1\/11 =2\/3-1\/11 =19\/33 希望能帮到你， 祝你学习进步，不理解请追问，理解请及时采纳！(*^__^*)
1\/1*3+1\/3*5+1\/3*7+1\/7*9+1\/9*11怎样简算?为什么?
应该是“1\/5*7”吧 1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11 =1\/2*(1-1\/3)+1\/2*(1\/3-1\/5)+1\/2*(1\/5-1\/7)+1\/2*(1\/7-1\/9)+1\/2*(1\/9-1\/11)=1\/2*(1-1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-1\/9+1\/9-1\/11)=1\/2*(1-1\/11)=1\/2*10\/11 =5\/11 ...
1\/1*3+1\/3*5+1\/5*7+...+1\/99*101=? (要简便)
解：令a1=1\/(1*3)、a2=1\/(3*5)、a3=1\/(5*7)、、an=1\/(99*101)。可得，an=1\/((2n-1)*(2n+1))当n=1，a1=1\/(1*3)，当n=2，a2=1\/(3*5)，则当n=50时，a50=1\/(99*101)所以1\/(1*3)+1\/(3*5)+1\/(5*7)+...+1\/(99*101)为数列an前50项的和S50。又an=1\/...
