最佳回答取４０分：幾題MATH=](都幾難下=[)

(1 - a^2)(1 - b^2) - 4ab
=1 - b^2 - a^2 + a^2b^2 - 4ab
=-(a^2 + 2ab + b^2 - a^2b^2 + 2ab - 1)
=-(a+b)^2(a^2b^2- 2ab - 1)

a^4 + 4.
=唔識sor

(a-b) (a+b)^3 - a^4 + b^4
=(a+b)^2(a^2-b^2) - a^4 + b^4
=a^4 + 2a^3b + a^2b^2 - a^2b^2 - 2ab^3 - b^4 - a^4 + b^4
=2a^3b - 2ab^3
=2ab(a^2 - b^2)
=2ab(a + b)(a - b)

9a^2 - b^2 + 3ab^2 -1
=(9a^2 - 1) + (3ab^2 - b^2)
=[(3a)^2 - (1)^2] + b^2(3a - 1)
=(3a - 1)(3a + 1) + b^2(3a-1)
=(3a - 1)(3a + 1 + b^2)

10(x + 1)^2 - 3(x + 1) - 18
=[2(x + 1) - 3][5(x + 1) + 6]
=(2x + 2 - 3)(5x + 5 + 6)
=(2x - 1)(5x + 11)

(m + 2n)^2 - 9mn
=m^2 + 4mn + 4n^2 - 9mn
=(m - 2n)^2 - mn
=之後唔識sor

2(2x - 1)^2 - 5(2x - 1) - 3
=[2(2x - 1) + 1][(2x - 1) - 3]
=(4x - 2 + 1)(2x - 1 - 3)
=(4x - 1)(2x - 4)
=2(4x - 1)(x - 2)

1/2 (x^2 +34x +4)
=唔識sor

Factorize x^2 - 2x -3

x^2 - 2x -3
=(x - 3)(x + 1)

Hence,factorize (a^2 + 2a)^2 - 2(a^2 + 2a) - 3

(a^2 + 2a)^2 - 2(a^2 + 2a) - 3
設(a^2 + 2a)是x
=x^2 - 2x -3
=(x - 3)(x + 1)
因為(a^2 + 2a)=x
所以=[(a^2 + 2a) - 3][(a^2 + 2a) + 1]
=(a^2 + 2 - 3)(a^2 + 2 + 1)
=(a + 3)(a - 1)(a + 1)^2
=(a + 3)(a + 1)(a^2 - b^2)←唔知仲駛唔駛呢步

2007-09-12 22:11:47 補充：
(m 2n)^2 - 9mn=m^2 4mn 4n^2 - 9mn=(m - 2n)^2 - mn=之後唔識sor應該係(m 2n)^2 - 9mn=m^2 4mn 4n^2 - 9mn=m^2 - 5mn 4n^2=(m - n)(m - 4n)

a^4+4
=a^4+4a^2+4-4a^2
=(a^2+2+2x)(a^2+2-2x)
(a-b) (a+b)^3 - a^4 + b^4
=(a-b) (a+b)^3 -( a^4 - b^4)
=(a-b)(a+b)[(a+b)^2+a^2+b^2)]
=(a+b)(a-b)(2a^2+2ab+2b^2)
=2(a+b)(a-b)(a^2+ab+b^2)

10(x+1)^2 - 3(x+1) - 18
=(2x+2+3)(5x+5-6)
=(2x+5)(5x-1)

2(2x-1)^2 - 5(2x-1) - 3
=2a^2-5a-3
=(a-3)(2a+1)
=(4x-1)(2x-4)
=2(x-2)(4x-1)

Factorize x^2 - 2x -3
=(x+1)(x-3)

Hence,factorize (x^2+2x)^2 - 2(x^2+2x) - 3
=(x^2+2x+1)(x^2+2x-3)
=(x-1)(x+3)(x+1)^2

2007-09-12 22:08:51 補充：
其他唔識計