十五道脱试计算题附加答案分数乘法

如题所述

  1、

  原式=(300-15/16)x1/15+(140-7/8)x1/7+(60-3/4)x1/3
    =20-1/16+20-1/8+20-1/4
    =60-(1/16+1/8+1/4)
    =60-7/16(通分)
    =59又9/16

 

  2、1+1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)×(1/2+1/3+1/4)=?

  解法:令1/2+1/3+1/4=a, 1/5=b
  原式=(1+a) ×(a+b)-(1+a+b)×a

    =a+a^2+b+ab-a-a^2-ab

    =b=1/5 。

     =5分之1

 

  3、1+1/3+1/3^2+1/3^3+……+1/3^100

  解答:

  令原式=A

  则3A=3+1+1/3+1/3^2+1/3^3+……+1/3^99=3+A-1/3^100

  解出方程:

  A=3/2-(2×3^100)^(-1)

 

  4、1/(1×2)+1/(2×3)+1/(3×4)+…+1/(99×100)

  解题方法:每个加号前或后面的数字变成两个数的减数再相加。

  1/(1×2)=1-1/2

  1/(2×3)=1/2-1/3

  1/(3×4)=1/3-1/4

  ……

  1/(99×100)=1/99-1/100

 

  5、1/(1×3)+1/(3×5)+1/(5×7)+…+1/(99×101)

  解题方法:将每个加号前或后面的数字提取同一系数后乘以两个数的减数再相加。

  1/(1×3)=(1/2)×(1-1/3)

  1/(3×5)= (1/2)×(1/3-1/5)

  1/(5×7)= (1/2)×(1/5-1/7)

  ……

  1/(99×101)= (1/2)×(1/99-1/101)

 

       6、1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+99/100

       å…ˆè½¬æ¢åˆ†æ•°ï¼Œç„¶åŽé€é¡¹ç›¸åŠ ï¼š

       5/6=1-(1/2-1/3)=1-1/2+1/3

       11/12=1-(1/3-1/4)=1-1/3+1/4

       19/20=1-(1/4-1/5)=1-1/4+1/5

       29/30=1-(1/5-1/6)= 1-1/5+1/6

       41/42=1-(1/6-1/7)=1-1/6+1/7

       55/56=1-(1/7-1/8)=1-1/7+1/8

       71/72=1-(1/8-1/9)= 1-1/8+1/9

       99/100=1-1/100

 

  7、1/(99×98×97)+1/(99×97×96)+1/(99×96×95)+…+1/(99×2×1)

  提取1/99公因式,再拆分分数,将每个加号前或后面的数字变成两个数的减数再相加。

  原式=(1/99)×[(1/97-1/98)+(1/96-1/97)+(1/95-1/96)+……+(1-1/2)]

    =(1/99)×(1-1/98)

    =(1/99)×(97/98)

    =97/[(100-1)×98]

    =97/(9800-98)

    =97/9702

 

  8、1×2/2+2×2/3+3×2/4+4×2/5+5×2/6+6×2/7+7×2/8+8×2/9+9×2/10

  =2×(1/2+2/3+3/4+4/5+5/6+6/7+7/8+8/9+9/10)

  =2×[(1-1/2)+(1-1/3)+(1-1/4)+…+(1-1/10)]

  =2×[9-(1/2+1/3+1/4+……+1/10)]

  =2×{9-[(1/2+1/3+1/6)+ (1/4+1/8)+(1/5+1/10)+1/7+1/9]}

  =2×[9-(1+3/8+3/10+1/7+1/9)]  分数分母化成2×4×5×7×9=2520进行通分

  =2×[8-1/2520×(3×5×7×9+3×4×7×9+2×4×5×9+2×4×5×7)

  =2×(8-2341/2520)

  =2×(7+179/2520)

  =14+179/1260

  =14又179/1260

 

  解题思路:

  将1变成分数放在最后面。第一个等式后括号里的每8个数分成一组,共能分248组,另外余下3个数,不参加分组。观察每组,加进16的和都是0。添加248个16进248组。

  1+1/1992×(1+2+3+4-5-6-7-8+9+10+11+12-13-14-15-16+17+18+19+20+…

     ã€€ã€€ã€€  +1979+1980-1981-1982-1983-1984+1985+1986)

  =1/1992×(1+2+3+4-5-6-7-8+9+10+11+12-13-14-15-16+17+18+19+20+…

        ã€€ã€€   +1979+1980-1981-1982-1983-1984+1985+1986+1992)

  =1/1992×[(16+1+2+3+4-5-6-7-8)+(16+9+10+11+12-13-14-15-16)

       ã€€ã€€ã€€+(16+17+18+19+20-21-22-23-24)+…

       ã€€ã€€ã€€+(16+1977+1978+1979+1980-1981-1982-1983-1984)+1985+1986+1992-248×16]

  =1/1992×(1985+1986+1992-3968)

  = 1995/1992

  =1又1/664

 

  =1+[(1+2) ×2/2]^(-1)+[(1+3) ×3/2] ^(-1) +[(1+4)×4/2]^(-1) +…+[(1+100) ×100/2]^(-1)

  =1+2×[(1/2)×(1/3)+ (1/3) × (1/4)+ (1/4) × (1/5)+ …+(1/100) × (1/101)]

  =1+2×(1/2-1/3+1/3-1/4+1/4-1/5+…+1/100-1/101)

  =1+2×(1/2-1/101)

  =1+1-2/101

  =1又99/101

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