1又1/3-7/12+9/20-11/30+13/42-15/56
=1+1/3-1/3-1/4+1/4+1/5-1/5-1/6+1/6+1/7-1/7-1/8
=1-1/8
=7/8本回答被网友采纳
一又1\/3-7\/12+9\/20-11\/30+13\/42-15\/56简便运算
原式=(1+1\/3)-(1\/3+1\/4)+(1\/4+1\/5)-(1\/5+1\/6)+(1\/6+1\/7)-(1\/7+1\/8)，打开括号，相邻两项相消，原式=1-1\/8=7\/8
1又3\/1-12\/7+20\/9-30\/11+42\/13-56\/15要按合理的方法计算
所以不知道你的原式是不是1+1\/3-7\/12+9\/20-11\/30+13\/42-15\/56。如果是上式的话，那么计算过程是：7\/12=4\/12+3\/12=1\/3+1\/49\/20=5\/20+4\/20=1\/4+1\/511\/30=6\/30+5\/30=1\/5+1\/613\/42=7\/42+6\/42=1\/6+1\/715\/56=8\/56+7\/56=1\/7+1\/8原式=1+1\/3-7\/12+9\/20-...
急1加3分之一减12分之7加20分之9减30分之11加42分之13减56分之15 用...
1-1\/3+7\/12-9\/20+11\/30-13\/42 =1-1\/3+(1\/3+1\/4)-(1\/4+1\/5)+(1\/5+1\/6)-(1\/6+1\/7)=1-1\/7 =6\/7
求1又1\/3 -7\/12+9\/20-11\/30+13\/42-15\/56的简便运算。
简单，注意到从第三项开始分子可以写成两个数的和，分母可以写成这两个数的积。因此用裂项法就行了 1+1\/3-7\/12+9\/20-11\/30+13\/42-15\/56 =1+1\/3-(3+4)\/3*4+(4+5)\/4*5-(5+6)\/5*6+(6+7)\/6*7-(7+8)\/7*8 =1+1\/3-1\/4-1\/3+1\/4+1\/5-1\/5-1\/6+1\/6+1\/7-...
简便计算 1又1\/3-7\/12+9\/20-11\/30+13\/42-15\/56
7\/12=1\/3+1\/4;9\/20=1\/4+1\/5...原式为1+1\/3-1\/3-1\/4+1\/4+1\/5-1\/5-1\/6+1\/6+1\/7-1\/7-1\/8=1-1\/8=7\/8
1又1\/3-7\/12+9\/20-11\/30+13\/42-15\/56的简便计算
1又1\/3-7\/12+9\/20-11\/30+13\/42-15\/56 =1+1\/3-1\/3-1\/4+1\/4+1\/5-1\/5-1\/6+1\/6+1\/7-1\/7-1\/8 =1-1\/8 =7\/8 或者1+1\/3-(1\/3+1\/4)+(1\/4+1\/5)-(1\/5+1\/6)+(1\/6+1\/7)-(1\/7+1\/8)=7\/8 可以这样哦！
1+1\/3-7\/12+9\/20-11\/30+13\/42-15\/56=? 简便计算,
1+1\/3-7\/12+9\/20-11\/30+13\/42-15\/56=1+(1\/3)-(1\/3)-(1\/4)+(1\/4)+(1\/5)-(1\/5)-(1\/6)+(1\/6)+(1\/7)-(1\/7)-(1\/8)=1-(1\/8)=7\/8
(技巧题)计算:1又1\/3-7\/12+9\/20-11\/30+13\/42-15\/56.
1又1\/3-7\/12+9\/20-11\/30+13\/42-15\/56 =(1+1\/3)-(1\/3+1\/4)+(1\/4+1\/5)-(1\/5+1\/6)+(1\/6+1\/7)-(1\/7+1\/8)=1+1\/3-1\/3-1\/4+1\/4+1\/5-1\/5-1\/6+1\/6+1\/7-1\/7-1\/8 =1-1\/8 =7\/8
1又3分之一减12分之7加20分之9减30分之11加42分之13减56分之15 简便...
主要用到这个知识：(2n+1)\/[nx(n+1)]=1\/n+1\/(n+1),如7\/12=1\/3+1\/4，所以：原式=1+1\/3-1\/3-1\/4+1\/4+1\/5-1\/5-1\/6+1\/6+1\/7-1\/7-1\/8...(将中间的相反数约掉)=1-1\/8 =7\/8
1又1\/3-7\/12+9\/20-11\/30+13\/42-15\/56简算
真简单 =1+1\/3-1\/3-1\/4+1\/4+1\/5-1\/5-1\/6+1\/6+1\/7-1\/7-1\/8 =1-1\/8 =7\/8
