想問
(x-3)²-4x²
要有步驟
唔該```
F.2 Factorization !!
2006-11-26 1:47 am
回答 (8)
2006-11-26 1:51 am
✔ 最佳答案
(x-3)²-4x² (x-3)²-(2x)²
=[(x-3)+2x][(x-3)-2x]
=(3x-3)(-x-3)
=-3(x-1)(x+3)
2006-11-25 17:55:06 補充：
a²-b²=(a+b)(a-b)
2006-11-26 2:21 am
因為(a+b)(a-b)
=a²-b²
所以(x-3)²-4x²
=[(x-3)+2x][(x-3)-2x]
=(3x-3)(-x-3)
=3(x-1)(-x-3)
=-3(x-1)(x+3)
=a²-b²
所以(x-3)²-4x²
=[(x-3)+2x][(x-3)-2x]
=(3x-3)(-x-3)
=3(x-1)(-x-3)
=-3(x-1)(x+3)
2006-11-26 1:58 am
(x-3)²-4x²
=(x-3)²-(2x)²
=[(x-3)+2x][(x-3)-2x]
=(x-3+2x)(x+3-2x)
=(3x-3)(-x+3)
=-(3x+3)(x-3)
=-3(x+3)(x-3)
=-3(x-3)²
=(x-3)²-(2x)²
=[(x-3)+2x][(x-3)-2x]
=(x-3+2x)(x+3-2x)
=(3x-3)(-x+3)
=-(3x+3)(x-3)
=-3(x+3)(x-3)
=-3(x-3)²
參考: me
2006-11-26 1:56 am
(x-3)²-4x²
solution
= x² - 3² - 6x - 4x²
= -9 - 6x - 3x² //
(錯唔關我事)
solution
= x² - 3² - 6x - 4x²
= -9 - 6x - 3x² //
(錯唔關我事)
參考: me
2006-11-26 1:54 am
(x-3)² - 4x²
=(x-3)²-(2x)²
= [(x-3)+(2x)] * [(x-3)-(2x)]
= (3x-3) (-3-x)
= - (3x-3) (3+x)
= - 3 (x-1)(3+x)
Reason:
x² - y²
=(x+y)(x-y)
2006-11-25 17:56:51 補充：
factorization 係得一個term咋，上面果個錯左
=(x-3)²-(2x)²
= [(x-3)+(2x)] * [(x-3)-(2x)]
= (3x-3) (-3-x)
= - (3x-3) (3+x)
= - 3 (x-1)(3+x)
Reason:
x² - y²
=(x+y)(x-y)
2006-11-25 17:56:51 補充：
factorization 係得一個term咋，上面果個錯左
2006-11-26 1:52 am
(x-3)²-4x²
=(x²-6x+9)-4x²
=-3x²-6x+9
=-3(x²+2x-3)
=-3(x+3)(x-1)
2006-11-25 17:53:17 補充：
上面2條友唔記得抽3出來係錯
=(x²-6x+9)-4x²
=-3x²-6x+9
=-3(x²+2x-3)
=-3(x+3)(x-1)
2006-11-25 17:53:17 補充：
上面2條友唔記得抽3出來係錯
2006-11-26 1:52 am
(x-3)²-4x²
= [(x-3)+2x][(x-3)-2x]
= (x-3+2x)(x-3-2x)
= (3x-3)(-x-3)
= -(x+3)(3x-3)
2006-11-25 17:53:16 補充：
= -3(x+3)(x-1)
= [(x-3)+2x][(x-3)-2x]
= (x-3+2x)(x-3-2x)
= (3x-3)(-x-3)
= -(x+3)(3x-3)
2006-11-25 17:53:16 補充：
= -3(x+3)(x-1)
收錄日期: 2021-04-18 20:58:57
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https://hk.answers.yahoo.com/question/index?qid=20061125000051KK03696