望采纳
=1又1/2×12+3又1/6×12+5又1/12×12+7又1/20×12+9又1/30×12
=18+38+61+84又3/5+108又2/5
=117+(84又3/5+108又2/5)
=117+193
=310本回答被提问者采纳
=18+38+61+423/5+544/5
=117+(423+544)/5
=117+193.4
=310.4
(9又1\/2+7又1\/6+5又1\/12+3又1\/20+1又1\/30)*12=
（9又1\/2+7又1\/6+5又1\/12+3又1\/20+1又1\/30）*12 =（9+7+5+3+1+1\/2+1\/6+1\/12+1\/20+1\/30）x12 =(25+1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/5-1\/6)x12 =(25+1\/2-1\/6)x12 =25x12+1\/2x12-1\/6x12 =25x4x3+6-2 =300+4 =304 ...
1又1\/2+3又1\/6+5又1\/12+7又1\/20+9又1\/42+11又1\/56+13又1\/72+15又1...
1又1\/2+3又1\/6+5又1\/12+7又1\/20+9又1\/42+11又1\/56+13又1\/72+15又1\/90 =(1+3+5+7+9+11+13+15)+(1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72+1\/90-1\/30)= 64+1\/(1*2 )+1\/(2*3 )+1\/(3*4) +1\/(4*5)+1\/(5*6)+1\/(6*7)+1\/(7*8)+1\/(8*...
1又1\/2+3又1\/6+5又1\/12+7又1\/20+9又1\/42+11又1\/56+13又1\/72+
=（1+3+..+15）+（1\/2+1\/2-1\/3+1\/3-1\/4+..+1\/9-1\/10）=64+1-1\/10=65-1\/10=64又9\/10
(1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10+1\/11+1\/12+1\/13+1\/14+1\/15...
所以将N分别取15到2 提取公因式1\/2 最后变成（1+3+4+...+14）\/2=52.5
1又1\/20+2又1\/30+3又1\/42+4又1\/56+5又1\/72+6又1\/90+7又1\/110=?
先把整数部分相加，是一个等差数列。即1+2+3+……+7=（1+7）*7*1\/2=28 再解决分数部分,1\/20=1\/4-1\/5 1\/30=1\/5-1\/6 以后同理,所以分数部分化简为1\/4-1\/5+1\/5-1\/6+1\/6-1\/7+……+1\/10-1\/11=1\/4-1\/11=7\/44 故结果为28又7\/44 ...
1+2又6分之1+3又12分之1+4又20分之1+5又30分之1+6又42分之1+7又56分...
原式=(1+2+3+...+9)+(1\/6 +1\/12+1\/20+...+1\/90)=45+[1\/(2×3)+1\/(3×4)+1\/(4×5)+...+1\/(9×10)]=45+1\/2 -1\/3 +1\/3 -1\/4+ 1\/4 -1\/5+...+1\/9 -1\/10 =45+ 1\/2 -1\/10 =45又5分之2 ...
1+3又1\/6+5又1\/12+7又1\/20+9又1\/30+11又1\/42+13又1\/56+15又1\/72+17...
最后一个数字如果是19又1\/110 就可以按下面的做法.思路：1\/6=1\/2-1\/3 、1\/12=1\/3-1\/4 、………1+3又1\/6+5又1\/12+7又1\/20+9又1\/30+11又1\/42+13又1\/56+15又1\/72+17又1\/90+19又1\/110 =（1+3+5+7+9+11+13+15+17+19）+（1\/6+1\/12+1\/20+1\/30+1\/42+1\/56...
求教解题过程! 1又1\/2-2又5\/6+3又1\/12-4又19\/20+5又1\/30-6又14\/42
比如7又1\/90=7+1\/9×10=7+1\/9-1\/10
1+3又六分之一加五又十二分之一加七又二十分之一加九分之三十分之一九...
原式=1+3+5+7+1\/6+1\/12+1\/21+3\/9+19\/10 =16+2\/12+1\/12+1\/21+1\/3+1+9\/10 =17+5\/20+18\/20+1\/21+7\/21 =18+3\/20+8\/21 =18+63\/420+160\/420 =18又223\/420
1+3又1\/6+5又1\/12+7又1\/20+9又1\/30+11又1\/42等于多少?
1+3又1\/6+5又1\/12+7又1\/20+9又1\/30+11又1\/42 =（1+3+5+7+9+11）+（1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/5-1\/6+1\/6-1\/7）=6×6+（1\/2-1\/7）=36又5\/14
