日行一善^^ f.4 maths

A.
f(x+3)
= (x+3)^2+3k
= x^2+6x+9+3k

g(x-3)
= (x-3)^2+k
= x^2-6x+9+k

B. f(x+3) - g(x-3)=12x+1
=> (x^2+6x+9+3k) - (x^2-6x+9+k) = 12x+1
=> 2k = 1
=> k = 1/2

it is given that f(x)= x^2+3k and g(x)= x^2+k
A...find f (x+3) and g(x-3)
f(x+3) = (x+3)^2 + 3k = x^2 + 6x + 9 + 3k
g(x-3) = (x-3)^2 + k = x^2 - 6x + 9 + k

B..if f (x+3) - g(x-3)= 12x+1 find the value of k

x^2 + 6x + 9 + 3k - (x^2 - 6x + 9 + k) = 12x + 1
x^2 + 6x + 9 + 3k - x^2 + 6x - 9 - k = 12x + 1
12x + 2k = 12x + 1
2k = 1
k = 1/2

2007-10-25 23:54:20 補充：
好多高手, 慢左, 不過我都做得幾清楚, 一步步投我一票, 唔該.... 你

f(x)= x^2+3k and g(x)= x^2+k
+++++++++++++++++++++++++++++++
A...
f (x+3)
= (x+3)^2+3k
= x^2+6x+9+3k

g(x-3)
= (x-3)^2+k
= x^2-6x+9+k
+++++++++++++++++++++++++++++++

B..
if f (x+3) - g(x-3)=12x+1, find the value of k
+++++++++++++++++++++++++++++++
f (x+3) - g(x-3)=12x+1
(x^2+6x+9+3k) - (x^2-6x+9+k)=12x+1
12x+2k=12x+1
2k=1
k = 1/2

有錯請指正!

2007-10-26 00:41:44 補充：
因為左右的 "12x" 互相消去了.