2、2又1/3x1又5/7x1又2/9x1又5/9x3/11x1又11/12x3又12/23
要写出过程!快!
å帮个å¿:3.75x4.23x36-125x0.423x3.8çç®ä¾¿è¿ç®
追ç3.75x4.23x36-125x0.423x3.8
=0.423ï¼3.75Ã360-125Ã3.8ï¼
=0.423ï¼5Ã7.5Ã36-5Ã25Ã3.8)
=0.423 (25Ã2Ã5Ã0.3Ã18-25Ã2Ã5Ã1.9)
=0.423Ã250(5.4-1.9)
=0.423Ã250Ã3.5
=0.10575Ã4Ã250Ã3.5
=105.75Ã3.5
=(100+5+0.7+0.05)Ã3.5
=350+17.5+2.45+0.175
=370.125
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简便计算:1、2\/12x14+2\/14x16+2\/16x18+2\/18x20+1\/20
1、2\/12x14+2\/14x16+2\/16x18+2\/18x20+1\/20 =1\/12-1\/14+1\/14-1\/16+1\/16-1\/18+1\/18-1\/20+1\/20 =1\/12 2.2又1\/3x1又5\/7x1又2\/9x1又5\/9x3\/11x1又11\/12x3又12\/23 =7\/3×12\/7×20\/9×14\/9×3\/11×23\/12×81\/23 =20×14\/11 =280\/11 =2...
2\/12x14+2\/14x16+2\/16x18+2\/18x20+1\/20 简算。
解答:2\/12x14+2\/14x16+2\/16x18+2\/18x20+1\/20 =1\/12-1\/14+1\/14-1\/16+1\/16-1\/18+1\/18-1\/20+1\/20 =1\/12
简便运算:2\/12x14+2\/12x16+2\/16x18+1\/20
=3\/24-1\/14+1\/32-1\/18+1\/20 =1\/8-1\/14-1\/18+1\/32+1\/20(这个好像题目有问题,不会这么复杂)2\/12x14+2\/14x16+2\/16x18+1\/20(把第二项的12改为14)=1\/12-114+1\/14-1\/16+1\/16-1\/18+1\/20 =1\/12-1\/18+1\/20 =1\/12-2\/18*20 =1\/12-1\/180 =15\/180-1\/180 =...
12x14\/2十14x16\/2十16\/1简便方法?
12x14\/2十14x16\/2十16\/1 =12×7+14×8+16 =84+112+16 =196+16 =212
2+4+6+……+16+18+20怎样用简便方法计算
2+4+6+……+16+18+20可以用等差数列来进行计算。即:Sn=a1*n+[n*(n-1)*d]\/2或Sn=[n*(a1+an)]\/2 2+4+6+……+16+18+20 ={(2+20)x20}÷2=22x10÷2=22x5=110
12×十四分之2+14x十六分之2+16分之一怎样简便运算算的多少
答:12×十四分之2+14x十六分之2+16分之一 =12*1\/7+29\/16 =12*16\/112+29*7\/112 =395\/112
2+4+6+?+18+20怎么简便计算
2+4+6+?+18+20简便计算方法如下:2+4+6+?+18+20=2+20)x(20÷2)÷2=22x10÷2=22x5=110。或者是2+4+6+?+18+20=(2+20)+(4+18)+??+(10+12)=22+22+22+22+22=22x5=110。第二个与第一个的差等于任一个与前一个的差,我们称为是等差数列,运用等差数列求和公式:Sn=(a1...
2\/8X10+2\/10X12+2\/12X14+2\/14X16
2\/8*10+2\/10*12+2\/12*14+2\/14*16 =1\/8-1\/10+1\/10-1\/12+1\/12-1\/14+1\/14-1\/16 =1\/8-1\/16 =2\/16-1\/16 =1\/16
谁能帮我解下方程:x\/20+(14-x)\/12=1
谁能帮我解下方程:x\/20+(14-x)\/12=1 我来答 1个回答 #热议# 为什么现在情景喜剧越来越少了?ysq512022 2013-08-18 · TA获得超过103个赞 知道小有建树答主 回答量:229 采纳率:0% 帮助的人:181万 我也去答题访问个人页 关注
1+2+3+4+5+...+20的简便算法?
经过仔细观察,第一个数字和最后一个数字相加等于21,第二个数字与倒数第二个数字相加等于21,以此类推,总共有20\/2个21,那么就可以得到算式为:(1+20)×20\/2 =21×10 =210 这就是这类算式的计算方法和步骤。