最好是一些能举一反三类的题目-查字典问答网
分类选择

来自罗建利的问题

  最好是一些能举一反三类的题目

  最好是一些能举一反三类的题目

1回答
2020-03-08 14:50
我要回答
请先登录
陆洲

  1.X和y是自然数,规定小x*y=4X-3y,如果5*a=8,那么a是几?

  a=(4*5-8)/3

  =4

  2.某班一次集合,请假人数是出席人数的1/9,中途又有一人请假离开,这样一来请假人是出席人数的3/22,那么这个班共有多少人?

  设第一次请假人为X,则出席人数为9X

  (9X-1)*(3/22)=x+1

  x=5

  所以:总人数为5+9*5=50(人)

  3.8又4分之3-0.35+(1又4分之1-6又20分之13)

  =8又3/4-7/20+1又1/4-6又13/20

  =(8又3/4+1又1/4)-(7/20-6又13/20)

  =3

  4.1/2*3+1/3*4+1/4*5.+1/49*50

  =1/2-1/3+1/3-1/4+1/4-1/5+.+1/49-1/50

  =1/2-1/50

  =12/25

  5.1/1*3+1/3*5+1/5*7+.+1/47*49

  =(1-1/3+1/3-1/5+1/5-1/7+.+1/47-1/49)*1/2

  =(1-1/49)*1/2

  =48/49*1/2

  =24/49

  6.1/2*5+1/5*8+1/8*11+.+1/20*23

  =(1/2-1/5+1/5-1/8+1/8-1/11+.+1/20-1/23)*1/3

  =(1/2-1/23)*1/3

  =21/46*1/3

  =7/46

  7.1又13+7/12-9/20+11/30-13/42

  =1又1/3+1/3+1/4-1/4-1/5+1/5+1/6-1/6-1/7

  =1又2/3-1/7

  =1又11/21

  8.2003/1*3+2002/3*5+2002/5*7+2002/7*9+2002/9*11

  =(1/1*3*2002+1/3*5*2002+1/5*7*2002+1/9*11*2002)*1/2

  =(1-1/3+1/3-1/5+1/5-1/7+1/9-1/11)*2002*1/2

  =(1-1/11)*2002*1/2

  =10/11*2002/2

  =9109.1/2+1/4+1/8+1/16+1/32

  =(1/2+1/4+1/8+1/16+1/32+1/32)-1/32

  =1-1/32

  =31/3210.1/12+1/20+1/30+1/42+1/56+1/72+1/90

  =1/3-1/4+1/4-1/5+1/5-1/6.+1/9-1/10

  =1/3-1/10

  =7/30

  11.1-1/4+1/20+1/30+1/42+1/56

  =1-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8

  =1-1/8

  =7/8

  12.1/1*5+1/5*9+1/9*13+.+1/55*59

  =(1-1/5+1/5-1/9+1/9-1/13+.+1/55-1/59)*1/4

  =(1-1/59)*1/4

  =29/118

  13.(1+1/3+1/5+1/7)*(1/3+1/5+1/7+1/9)-(1+1/3+1/5+1/7+1/9)+(1/3+1/5+1/7)

  设(1+1/3+1/5+1/7)为a,(1/3+1/5+1/7)为b

  a*(b+1/9)-(a+1/9)*b

  =ab+1/9a-ab-1/9b

  =1/9*(a+b)

  =1/9

  14.5/14*5/6-7/12*5/14+9/20*5/14

  =5/14*(5/6-7/12+9/20)

  =5/14*[1/2+1/3-(1/3+1/4)+1/4+1/5]

  =5/14*(1/2+1/3-1/3-1/4+1/4+1/5)

  =5/14*7/10

  =1/4

  15.1/3+1/15+1/35+1/63+1/99

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

  =(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)*1/2

  =(1-1/11)*1/2

  =5/11

2020-03-08 14:52:32

最新问答

推荐文章

猜你喜欢

附近的人在看

推荐阅读

拓展阅读

  • 大家都在看
  • 小编推荐
  • 猜你喜欢
  •