幾條數學難題...

1.(9÷8)÷(8÷7)÷(7÷6)÷(6÷5)÷(5÷4)÷(4÷3)÷(3÷2)÷(2÷1)
=9/8 x 7/8 x 6/7 x 5/6 x 4/5 x3/4 x 2/3 x 1/2
=9/(8x8)
=9/64

2. 設已考了x 次測驗
則x次測驗的總分為 84x
這一次是100分的話平均分成為
(84x+100)/(x+1)=86
84x+100=86x+86
100-86=86x-84x
14=2x
x=7
所以這次是第x+1次=第8次測驗

3.設排球的數量是x
藍球的數量是y
足球的數量是z
x+y+z=203 ----(1)
x=2y ----(2)
z=x-2 ---(3)

代表(2)入(3)
z=2y-2 ---(4)

代表(2),(4) 入 (1)
2y+y+2y-2=203
5y=203+2
y=205/5
y=41 ---(5)
代(5)入(2)
x=82
代(5)入(4)
z=80

所以有籃球41個、足球80個和排球82個。

1.(9÷8)÷(8÷7)÷(7÷6)÷(6÷5)÷(5÷4)÷(4÷3)÷(3÷2)÷(2÷1)
=9/8÷8/7÷7/6÷6/5÷5/4÷4/3÷3/2÷2/1
=9/8x7/8x6/7x5/6x4/5x3/4x2/3x1/2
=9/64
2.我不知道要求上次還是今次的，就當是今次。
設上次是第b次的測驗
(84b+100)/(b+1)=86
84b+100=86(b+1)
84b+100=86b+86
84b+14=86b
14=2b
7=b
b=7
答:今次是第(7+1)=8次的測驗
3.設籃球有a個，則排球有2a個，足球有(2a-2)個
a+2a+(2a-2)=203
5a-2=203
5a=205
a=41
答:籃球有41個，排球有(41x2)=82個，足球有(82-2)=80個

1. 9/8 x7/8 x6/7 x5/6 x 4/5 x3/4 x2/3 x1/2 =9/64

2. let the number of tests he has done be Y

84 x Y+ 100 = (Y+1) x 86

84Y +100 = 86Y +86

14= 2Y

Y=7

therefore, the last time is the seventh time.

3. let the number of basketball is X.

then the number of volleyball is 2X

the number of football is 2X-2

X + 2X + (2X-2) =203

5X -2 = 203

5X = 205

X=41

therefore, there is 41 basketball, 82 volleyball and 80 football.

這是否奧數?

1.9/8÷8/7÷7/6÷6/5÷5/4÷4/3÷3/2÷2
=9/64

2.第5次

3.i don't know

hope can help you