中4 數學 聯變 急急

1.
a ∝ b/√c
所以 a = kb/√c，k 為常數

當 b = bo 及 c = co, a = ao:
ao = kbo/√co

當 b = bo(1 + 30%) 及 c = co(1 - 75%):
a = kbo(1 + 30%)/√[co(1 - 75%)]
a = 1.3kbo/co√0.25
a = 2.6kbo/co
a = 2.6ao

a 的百分變化
= [(2.6ao - ao)/ao] x 100%
= +160%
（增加 160%）


2.
(a)
y ∝ (x + 3)(x - 2)
y = k(x + 3)(x - 2)，k 為常數

當 x = 5，y = 12 :
12 = k(5 + 3)(5 - 2)
k = 0.5

所以，y = 0.5(x + 3)(x - 2)

(b)
當 y = 25 :
25 = 0.5(x + 3)(x - 2)
50 = x² + x - 6
x² + x - 56 = 0
(x + 8)(x - 7) = 0
x = -8 或 x = 7


3.
(a)
m ∝ (n + a)(n - 1)
所以 m = k(n + a)(n - 1)，k 為常數

當 n = 3，m = 40
40 = k(3 + a)(3 - 1)
k(3 + a) = 20 ...... [1]

當 n = 5，m = 112
112 = k(5 + a)(5 - 1)
k(5 + a) = 28 ...... [2]

[1]/[2] :
(3 + a)/(5 + a) = 20/28
(3 + a)/(5 + a) = 5/7
21 + 7a = 25 + 5a
2a = 4
a = 2

(b)
把 a = 2 代入 [1] :
k(3 + 2) = 20
k = 4

所以，m = 4(n + 2)(n - 1)

(c)
當 m = 72 :
72 = 4(n + 2)(n - 1)
n² + n - 2 = 18
n² + n - 20 = 0
(n + 5)(n - 4) = 0
n = -5 或 n = 4


4.
s ∝ t³
s = kt³，k 為常數

當 t = 4 :
s = k(4)³
s = 64k

當 t = 4(1 + 50%) :
s = k[4(1 + 50%)]³
s = k(6)³
s = 216k

s 的改變百分比
= [(216k - 64k)/64k] x 100%
= +237.5%
（由 100% 增至 337.5%）


5.
題目中是 s 隨 x² 甚麼？題目中也一直沒有 y。


6.
(a)
題目可能是 z 部份是常數，部份隨 xy 而反變。
z = k1 + (k2/xy)
k1 及 k2 為常數

當 x = 6 及 y = 7，z = 2 :
2 = k1 + (k2/42)
2 * 42 = [k1 + k2(1/42)] * 42
42k1 + k2 = 84 ...... [1]

當 x = 2 及 y = 3，z = 5 :
5 = k1 + k2/6)
5 * 6 = [k1 + (k2/6)] * 6
6k1 + k2 = 30 ...... [2]

[1] - [2] :
36k1 = 54
k1 = 3/2

把 k1 = 3/2 代入 [2] :
6(3/2) + k2 = 30
k2 = 21

所以，z = (3/2) + (21/xy)

(b)
當 y = 2 及 z = -2 :
-2 = (3/2) + (21/2x)
21/2x = -7/2
x = -3

(c)
當 z = 3/xy: xy = 3/z
所以 z = (3/2)+ [21/(3/z)]
z = (3/2) + 7z
6z = -3/2
z = -1/4

2012-03-04 01:14:49 補充：
這題沿用以往方法便可。
a.
L = km³/√n
把 m, n, L 的值代入上式，k = 1/4
故 L = m³/4√n

b.
把 n 和 L 的值代入上式，m = 10