中四數學~ 20分

1.
9^(1/h)=99
(1/h)log(9) = log(99)
1/h = log(99)/log(9)
h = log(9)/log(99)

11^(1/k) =99
(1/k)log(11) = log (99)
1/k = log(99)/log(11)
k = log(11)/log(99)

h+k
= log(9)/log(99) + log(11)/log(99)
= [log(9) + log(11)]/log(99)
= log(9x11)/log(99)]
= log(99)/log(99)
= 1

2. 已知f(x)=100^logx
a. f(1)
= 100^log (1)
= 10^2log(1)
= 10^0
= 1

b. f(2)+f(3)
= 100^log(2) + 100^log(3)
= 10^2log(2) + 10^2log(3)
= 10^log(4) + 10log^(9)
= 4 + 9
= 13

3.
log【(√179 - 13)^20 (√179 + 13)^20】
= log [(179 - 169)^20]
= log (10^20)
= 20

4.
(k-1)^3 > 2005
3log(k - 1) > log (2005)
log (k - 1) > log (2005) / 3
k - 1 > 10 ^ [log (2005) / 3]
k > 10 ^ [log (2005) / 3] + 1
k > 13.6
最小整數 k 的值 = 14.

5.
a.
2^x = 10^z
z = x log (2)
x = z / log (2)

5^y = 10^z
z = y log (5)
y = z / log (5)

b.
1/x + 1/y
= 1/[z / log (2)] + 1/[z / log(5)]
= log (2)/z + log (5)/z
= [log(2)+log(5)] / z
= log (2 x 5) / z
= 1/z

6. 若x^(1/4) + x^(-1/4) =4
a.
√x + 1/√x
= [(x^1/4 + x^-1/4)^2] - 2
= (4^2) - 2
= 16 - 2
= 14
b. x + 1/x
= [(√x + 1/√x)^2] - 2
= (14^2) - 2
= 196 - 2
= 194

有錯請指正!

2007-10-25 21:23:04 補充：
無問題!√x = x^(1/2) = [x^(1/4)]^21/√x = x^(-1/2) = [x^(-1/4)]^22 (√x) (1/√x) = 2因為 [(x^1/4 x^(-1/4)]^2 = √x 1/√x 2 [(a b)^2 = a^2 b^2 2ab]所以, √x 1/√x = √x 1/√x 2 - 2= [(x^1/4 x^(-1/4)]^2 - 2其餘, 照著上面的類似方法去計.如果仍然不明白, 歡迎再提問! 謝謝!

2007-10-25 21:24:06 補充：
無問題!√x = x^(1/2) = [x^(1/4)]^21/√x = x^(-1/2) = [x^(-1/4)]^22 (√x) (1/√x) = 2因為 [(x^1/4 + x^(-1/4)]^2 = √x + 1/√x + 2 [(a+b)^2 = a^2+b^2+2ab]所以, √x + 1/√x = √x + 1/√x + 2 - 2= [(x^1/4 + x^(-1/4)]^2 - 2其餘, 照著上面的類似方法去計.如果仍然不明白, 歡迎再提問! 謝謝!

2007-10-26 10:54:13 補充：
其實我之前比較趕, 所以沒有寫清楚.b. x = [x^(1/2)]^2 = (√x)^21/x = x^(-1) = [x^(-1/2)]^2 = (1/√x)^22 (x) (1/x) = 2因為 [√x + (1/√x)]^2 = x + 1/x + 2 用公式[(a+b)^2 = a^2+b^2+2ab]所以, x + 1/x = x + 1/x + 2 - 2= [√x + (1/√x)]^2 - 2

2007-10-26 10:55:16 補充：
其實, 竅門是用公式[(a+b)^2 = a^2+b^2+2ab], 之後b. part用a. part 的結果.有興趣不妨加多一part, x^2 + 1/(x^2) = ? [Ans. 37634] 如果你計到, 你以後見到類似題目都應該計到!如果仍然不明白, 歡迎再提問! 謝謝!