初中數學問題+排列與組合求解答!要詳細步驟PLZ

Q1)
[a^(n+4) - a^(3n)] / [a^(3n+3) - a^(n+7)]
= [a^(n+4) - a^(3n)] / [a^3•a^(3n) - a^3•a^(n+4)]
= -[a^(3n) - a^(n+4)] / (a^3)[ a^(3n) - a^(n+4)]
= -1/a^3


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Q2)
[2^3 • (-2)^(2n - 1)] / 2(-2)^4n
= [-(-2)^3 • (-2)^(2n - 1)] / 2(2)^4n
= -(-2)^(3+2n-1) / 2^(4n+1)
= -(-2)^(2n+2) / 2^(4n+1)
= -2^(2n+2) / 2^(4n+1)
= -2^(2n+2-4n-1)
= -2^[-(2n-1)]
= -1/2^(2n-1)


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Q3)
Let y cm the length of each side of the square.
Let h cm be the length of the original height.

(1/3) x 5 x 9 x h = (1/3) x y² x (1.25h)
1.25y = 45
y = 36
y = 6 or y = -6 (rejected)

The length of each side of the square = 6 cm


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Q4)
The frustum is made by cutting off a small pyramid from the original pyramid.
The upper base is the base of the small pyramid, while the lower base is thebase of the original pyramid.

Height of the small pyramid
= 126 - 42 cm
= 84

Volume of the frustum
= (Volume of the original pyramid) - (Volume of the small pyramid)
= (1/3) x 30² x 126 - (1/3) x 20² x 84 cm³
= 26600 cm³


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Q5)
(a)
設原有酒精的體積為 4 單位，則原有水的體積為 1 單位。

新混合物中，酒精體積 : 水體積
= 4(1 - 15%) : 1(1 + 30%)
= 3.4 : 1.3
= 34 : 13

(b)
總體積的百分變化
= {[(3.4 + 1.3) - (4 + 1)] / (4 + 1)} x 100%
= -(0.3/5) x 100%
= -6%
(減少 6%)


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Q6)
設高應減少 R%。

(1 + 25%) x (1 - R%) = 1
1.25(1 - R%) = 1
1 - R% = 0.8
R% = 0.2
R = 20

高應減少 20%。


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Q7)
(a)
在哥哥送棒棒糖給弟弟後，弟弟的棒棒糖數目
= 20 x [3/(2+3)] 枝
= 20 x (3/5) 枝
= 12 枝

(b)
設哥哥原有棒棒糖 n 枝。
則弟弟原有棒棒糖 (20 - n) 枝。

(20 - n) + (1/3)n = 12
20 - n + (1/3)n = 12
(2/3)n = 8
n = 12

在哥哥送棒棒糖給弟弟前，哥哥的棒棒糖數目 = 12 枝


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Q8)
(a)
x³ - x²y + xy²
= x(x² - xy + y²)

(b)
x²y - xy² + y³
= y(x² - xy + y²)

(c)
x³ + y³
= (x³ - x²y + xy²) + (x²y - xy² + y³)
= x(x² - xy + y²) + y(x² - xy + y²)
= (x + y)(x² - xy + y²)


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Q9)
在 11 個字母中抽出 3 個不同的字母，不同的組合數目
= 11C3 個
= 11! / 3!8!
= 165 個


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Q10)
牠們在相同月份出生的概率
= 1 x (1/12)
= 1/12

牠們在不同月份出生的概率
= 1 - (1/12)
= 11/12