幾題中三的數學問題，唔該教教小弟。請列式。

1)
(x+1)(x+2)(x+3) = 15(x+2)
(x+1)(x+2)(x+3) - 15(x+2) = 0
(x+2) [ (x+1)(x+3) - 15 ] = 0
(x+2) [ x^2 + 4x + 3 - 15 ] = 0
(x+2)(x^2 + 4x - 12) = 0
(x+2)(x+6)(x-2) = 0
x = 2 or x = -2 or x = -6

2)
x^3 - x = 6(x-1)
x(x^2-1) - 6(x-1) = 0
x(x+1)(x-1) - 6(x-1) = 0
(x-1) [ x(x+1) - 6 ] = 0
(x-1)(x^2 + x - 6) = 0
(x-1)(x+3)(x-2) = 0
x = 1 or x = -3 or x = 2

3)
kx(x - 1/k) = 2/k
kx^2 - x = 2/k
(kx)^2 - kx - 2 = 0
(kx-2)(kx+1) = 0
kx = 2 or kx = -1
x = 2/k or x = -1/k

4)
x^2 + 2ax = c^2 + 2ac
x^2 + 2ax - c^2 - 2ac = 0
(x^2 - c^2) + (2ax - 2ac) = 0
(x+c)(x-c) + 2a(x-c) = 0
(x-c) [ (x+c) + 2a ] = 0
(x-c)(x+c+2a) = 0
x = c or x = -2a-c

5)
5x^2 - 4nx = 5e^2 - 4en
5x^2 - 4nx - 5e^2 + 4ne = 0
(5x^2 - 5e^2) - (4nx - 4ne) = 0
5(x+e)(x-e) - 4n(x-e) = 0
(x-e) [ 5(x+e) - 4n ] = 0
(x-e)(5x+5e-4n) = 0
x = e or 5x = 4n - 5e
x = e or x = (4n-5e)/5

6)
x^2 - 2x - (a^2 - 2a) = 0
x^2 - 2x - a^2 + 2a = 0
(x^2 - a^2) - (2x - 2a) = 0
(x+a)(x-a) - 2(x-a) = 0
(x-a)(x+a-2) = 0
x = a or x = 2-a

7)
x^2 - 2px - (q^2 - p^2) = 0
(x^2 - 2px + p^2) - q^2 = 0
(x-p)^2 - q^2 = 0
(x-p+q)(x-p-q) = 0
x = p-q or x = p+q

8)
x^2(a+2b) = a^2(x+2b)
ax^2 + 2bx^2 = a^2x + 2a^2b
ax^2 - a^2x + 2bx^2 - 2a^2b = 0
ax(x-a) + 2b(x^2 - a^2) = 0
ax(x-a) + 2b(x+a)(x-a) = 0
(x-a) [ ax + 2b(x+a) ] = 0
(x-a)(ax+2bx+2ab) = 0
x = a or ax+2bx+2ab = 0
x = a or x(a+2b) = -2ab
x = a or x = -2ab/(a+2b)

9)
x^2(a-3b) = b^2(a+3x)
ax^2 - 3bx^2 = ab^2 + 3b^2x
(ax^2 - ab^2) - (3bx^2 + 3b^2x) = 0
a(x+b)(x-b) - 3bx(x+b) = 0
(x+b) [ a(x-b) - 3bx) ] = 0
(x+b)(ax-b-3bx) = 0
x+b = 0 or ax-3bx-b = 0
x=-b or x(a-3b) = b
x = -b or x = b/(a-3b)

10)
設 a 為開方 p
a^2 - a - 6 = 0
(a-3)(a+2) = 0
a = 3 or a = -2
因此
開方p = 3 or 開方p = -2 (rej)
p = 9

11)
設 a 為開方 q
a^2 - 3a - 10 = 0
(a-5)(a+2) = 0
a = 5 or a = -2
開方q = 5 or 開方q = -2 (rej)
q = 25

that is no need to be rejected

10.p-(開方p)-6=0
(√p-3)(√p+2)=0
p=9 or p=-2^2=4

11.q-3(開方q)-10=0
(√q-5)(√q+2)=0
q=25 or q=-2^2=4

2007-11-06 18:58:45 補充：
answer will be rejected when x^2= -yor the denominator is 0