六條數學難題

1.
(a)
這兩位數的值
= 10x + (x + 2)
= 11x + 2

(b)
11x + 2 = 3x(x + 2)
11x + 2 = 3x² + 6x
3x² - 5x - 2 = 0
(3x + 1)(x - 2) = 0
x = -1/3(捨去) 或 x = 2

該數為 24。


=====
2.
(a)
利潤
= $(y - 20)(3600 - 60y)

(b)
(y - 20)(3600 - 60y)
= -60(y - 20)(y - 60)
= -60(y² - 80y + 1200)
= -60(y² - 80y) - 72000
= -60(y² - 80y + 40²) + 60*40² - 72000
= 24000 - 60(y - 40)²

就任何實數 y，(y - 40)² ≥ 0
因此，利潤 = $[24000 - 60(y - 40)²] ≤$24000
最高利潤 = $24000


=====
3.
(2a^3 *b^-2)^-2
= 2^-2 * a^-6 * b^4
= 4^-1 * a^-6 * b^4
= b^4/(4a^6)


=====
4.
[3^(n-1) + 3^(n+1)] / 3n
= [3^(n-1) + 3^2*3^(n-1)] / 3n
= (1 + 9)*3^(n-1) / 3n
= 10*3^(n-2)


=====
5.
3log4 + log2x = log768
log4^3 + log2x = log768
log64 + log2x = log768
log[64(2x)] = log768
log128x = log768
128x = 768
x = 6


=====
6.
1 - cosθ = sin²θ
1 - cosθ = 1 - cos²θ
cos²θ - cosθ = 0
cosθ(cosθ - 1) = 0
cosθ = 0 或cosθ = 1
θ = 90°, 270° 或θ =0°, 360°
θ = 0°, 90°, 270°, 360°

2012-09-09 13:58:48 補充：
4.
題目應是： [3^(n-1) + 3^(n+1)] / 3^n

解法如下：
[3^(n-1) + 3^(n+1)] / 3^n
= [3^(n-1) + 3^2*3^(n-1)] / 3*3^(n-1)
= (1+9)*3^(n-1) / 3*3^(n-1)
= 10/3