1/1*3+1/3*5+1/5*7+1/7*9+1/9*11+1/11*13

1\/1×3+1\/3×5+1\/5×7+1\/7×9+1\/9×11+1\/11×13的简便运算
=1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11+1\/11*13 =1\/2*[(1-1\/3)+(1\/3-1\/5)+(1\/5-1\/7)+(1\/7-1\/9)+(1\/9-1\/11)+(1\/11-1\/13)} =1\/2*[1-1\/13]=6\/13

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

1\/1x3+1\/3x5+1\/5x7+1\/7x9+1\/9x11=多少
解析 因为，1\/[(2n－1)(2n+1)]=1\/2[1\/(2n－1)－1\/(2n+1)]所以，原式=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\/7－1\/9+1\/9－1\/11)=1\/2(1－1\/11)=1\/2×10\/11 =5\/11 本...

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...

1\/1*3+1\/3*5+1\/5*7+1\/7*9+19*11+...+1\/1995*1997+1\/1997*1999的简便算 ...
观察数字，你会发现这样的规律：1\/1*3=（1-1\/3）*0.5 1\/3*5=（1\/3-1\/5）*0.5 1\/5*7=（1\/5-1\/9）*0.5 1\/7*9=（1\/7-1\/9）*0.5 ...那么1\/1*3+1\/3*5+1\/5*7+1\/7*9+19*11+...+1\/1995*1997+1\/1997*1999 =（（1-1\/3）+（1\/3-1\/5）+（1\/5-1\/9...

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！(*^__^*)

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
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)=5\/11,,,如果后面还有依次类推，就可以得出答案，新手，求采纳啊

1\/1×3+1\/3×5+1\/5×7+1\/7×9+1\/9×11+……+1\/19×21
1\/1×3＋1\/3×5＋1\/5×7＋1\/7×9＋1\/9×11＋……＋1\/19×21 =(1\/2)(1\/1-1\/3+1\/3-1\/5+...+1\/19-1\/21)=(1\/2)(1-1\/21)=10\/21 把每一个都拆成两项相减 比如1\/1×3=(1\/1-1\/3)\/2

1\/1×3×5+1\/3×5×7+1\/5×7×9+1\/7×9×11+1\/9×11×13+1\/
分子和分母同时乘(1×3×5×7×9×11×13×15),然后再除这个数就可以了，把多次的除法变成多次乘法和仅仅一次的除法。