f.3 maths...

x2-y2+4x+8y-12 點計呀??
= x2 + 4x + 4 – y2 + 8y - 16
= (x + 2)2 – (y – 4)2
= (x + 2 – y + 4)(x + 2 + y – 4)
= (x – y + 6)(x + y – 2)

同埋呢我想問
(4x-2y-3)(4x+2y+3) 同 (4x+2y-3)(4x-2y+3)有咩分別呀??
兩個數式中只有一個正負號不同，整個式已經是不同。
(4x-2y-3)(4x+2y+3) 這式未分解之前為
= (4x)2 – (2y + 3)2
而
(4x+2y-3)(4x-2y+3)
= (4x)2 – (2y – 3)2

x^2-y^2+4x+8y-12
=x^2+4x-y^2-8y-12
=x^2+4x-(y^2+8y+12)
=x^2+4x-(y+2)(y+6)
=(x-y-2)(x+y+6)

(4x-2y-3)(4x+2y+3)
=16x^2+8xy+12x-8xy-4y^2-6y-12x+6y-9
=16x^2-4y^2-9

(4x+2y-3)(4x-2y+3)
=16x^2-8xy+12x+8xy-4y^2+6y-12x+6y-9
=16x^2-4y^2+12y-9

因為16x^2-4y^2-9=/=16x^2-4y^2+12y-9

所以(4x-2y-3)(4x+2y+3)=/=(4x+2y-3)(4x-2y+3)

2006-10-22 22:13:21 補充：
sor應該係咁x^2-y^2+4x+8y-12=x^2+4x-y^2+8y-12=x^2+4x-(y^2-8y+12)=x^2+4x-(y-2)(y-6)=(x+y-2)(x-y-6)