(x-3)二次減4x二次
要詳細
f.2 maths一小條
2007-11-15 4:11 am
回答 (6)
2007-11-15 4:15 am
✔ 最佳答案
^2 = 二次(x-3)^2-4x^2
=x^2-6x+9-4x^2
=-3x^2-6x+9
=-3(x^2+2x-3)
=-3(x-1)(x+3)
2007-11-14 22:17:45 補充：
Q ) 點解要做 (-x - 3)(3x - 3)= -3(x 3)(x - 1) ??A ) 因為 (x^2 2x-3) 仲可以繼續 factorize/ simplify 佢 用 cross method 就可以知道 (x^2 2x-3) )= (x 3)(x - 1)明唔明白？=]
2007-11-14 22:20:43 補充：
我上面 expand 左佢都 work或者可以用 A^2 - B^2 = (A+B)(A-B) 呢條式(x-3)^2-4x^2=(x-3)^2 - (2x)^2= (x-3+2x) ( x-3-2x)= (3x-3)(-x-3)= -3(x+3)(x-1)
2007-11-15 4:31 am
(x-3)^2-4x^2
=(x^2-6x+9)-4x^2
=x^2-6x+9-4x^2
= -3x^2-6x+9
=(x^2-6x+9)-4x^2
=x^2-6x+9-4x^2
= -3x^2-6x+9
2007-11-15 4:18 am
Method I)
(x-3)^2 - 4x^2
=(x^2-6x+9)-4x^2
=-3x^2-6x+9
=-3(x^2+2x-3)
=-3(x+3)(x-1)
Method II)
(x-3)^2 - 4x^2
=(x-3)^2 - (2x)^2
=[(x-3)+2x][(x-3)-2x] [A^2-B^2=(A+B)(A-B)]
=(3x-3)(-x-3)
=-3(x-1)(x+3)
(x-3)^2 - 4x^2
=(x^2-6x+9)-4x^2
=-3x^2-6x+9
=-3(x^2+2x-3)
=-3(x+3)(x-1)
Method II)
(x-3)^2 - 4x^2
=(x-3)^2 - (2x)^2
=[(x-3)+2x][(x-3)-2x] [A^2-B^2=(A+B)(A-B)]
=(3x-3)(-x-3)
=-3(x-1)(x+3)
2007-11-15 4:16 am
(x-3)^2 - 4x^2
= (x-3)^2 - (2x)^2
= (x - 3 - 2x)(x - 3 + 2x)
= (-x - 3)(3x - 3)
= -3(x + 3)(x - 1)
2007-11-14 20:18:47 補充：
oh, latebut please give me one vote
= (x-3)^2 - (2x)^2
= (x - 3 - 2x)(x - 3 + 2x)
= (-x - 3)(3x - 3)
= -3(x + 3)(x - 1)
2007-11-14 20:18:47 補充：
oh, latebut please give me one vote
2007-11-15 4:16 am
(x-3)(x-3)-(4x)(4x)
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https://hk.answers.yahoo.com/question/index?qid=20071114000051KK03354