我想問:
1) 3(x-y)-6a(y-x)
2) (m-n)^2+n(n-m)
點計？求steps!!!
Factorize these questions!!! 求steps!!!
2006-11-22 2:48 am
回答 (6)
2006-11-22 2:58 am
✔ 最佳答案
1. 3(x - y) - 6a(y - x)= 3(x - y) + 6a(x - y) < remember that -(x - y) = y - x
= (3 + 6a)(x - y) < x- y is the H.C.F
= 3(1 + 2a)(x - y)
2. (m - n)^2 + n(n - m)
= (m - n)(m - n) + n(n - m)
= (m - n)(m - n) - n(m - n) < remember that -(x - y) = y - x
= (m - n - n)(m - n)
= (m - 2n)(m - n)
參考: me
2006-11-22 3:04 am
1) 3(x-y)-6a(y-x)
=3x-3y-6ay+6ax
=(3x+6ax)-(3y+6ay)
=3x(1+2a)-3y(1+2a)
=(1+2a)(3x-3y)
=3x-3y-6ay+6ax
=(3x+6ax)-(3y+6ay)
=3x(1+2a)-3y(1+2a)
=(1+2a)(3x-3y)
2006-11-22 3:02 am
1)Tips:
3(x-y)-6ay+6ax
=3(x-y)+6ax-6ay
=3(x-y)+6a(x-y)
=(3+6a)(x-y)
3(x-y)-6ay+6ax
=3(x-y)+6ax-6ay
=3(x-y)+6a(x-y)
=(3+6a)(x-y)
2006-11-22 3:01 am
1. 3(x - y) - 6a(y - x)
= 3(x - y) + 6a(x - y) ← -(x - y) = y - x
= (3 + 6a)(x - y)
= 3(1 + 2a)(x - y)
2. (m - n)^2 + n(n - m)
= (m - n)(m - n) + n(n - m)
= (m - n)(m - n) - n(m - n) ← -(x - y) = y - x
= (m - n - n)(m - n)
= (m - 2n)(m - n)
= 3(x - y) + 6a(x - y) ← -(x - y) = y - x
= (3 + 6a)(x - y)
= 3(1 + 2a)(x - y)
2. (m - n)^2 + n(n - m)
= (m - n)(m - n) + n(n - m)
= (m - n)(m - n) - n(m - n) ← -(x - y) = y - x
= (m - n - n)(m - n)
= (m - 2n)(m - n)
2006-11-22 2:59 am
3(x-y)-6a(y-x)
=3(x-y)+6a(x-y)
=(3+6a)(x-y)
(m-n)^2+n(n-m)
=(m-n)^2-n(m-n)
=(m-n)(m-n)-n(m-n)
=(1-n)(m-n)^2
I hope I'm correct
2006-11-21 19:01:07 補充：
(m-n)^2 n(n-m)=(m-n)(m-n)-n(m-n)=(m-n)(m-n-n)=(m-n)(m-2n) sorry , I do the second one wrong
=3(x-y)+6a(x-y)
=(3+6a)(x-y)
(m-n)^2+n(n-m)
=(m-n)^2-n(m-n)
=(m-n)(m-n)-n(m-n)
=(1-n)(m-n)^2
I hope I'm correct
2006-11-21 19:01:07 補充：
(m-n)^2 n(n-m)=(m-n)(m-n)-n(m-n)=(m-n)(m-n-n)=(m-n)(m-2n) sorry , I do the second one wrong
2006-11-22 2:57 am
1) 3(x-y)-6a(y-x)
=3(x-y) + 6a(x-y)
=(3+6a)(x-y)
2) (m-n)^2+n(n-m)
=(m-n)(m-n)-n(m-n)
=(m-n)(m-n-n)
=(m-n)(m-2n)
2006-11-22 22:03:42 補充：
3) -ab(rx-s)-ab^2(s-rx)^2=ab(s-rx) - ab^2(s-rx)^2=(s-rx)[ab-ab^2(s-rx)]=(s-rx){(ab)[1-b(s-rx)]}=(s-rx)(ab)[1-b(s-rx)]
=3(x-y) + 6a(x-y)
=(3+6a)(x-y)
2) (m-n)^2+n(n-m)
=(m-n)(m-n)-n(m-n)
=(m-n)(m-n-n)
=(m-n)(m-2n)
2006-11-22 22:03:42 補充：
3) -ab(rx-s)-ab^2(s-rx)^2=ab(s-rx) - ab^2(s-rx)^2=(s-rx)[ab-ab^2(s-rx)]=(s-rx){(ab)[1-b(s-rx)]}=(s-rx)(ab)[1-b(s-rx)]
收錄日期: 2021-04-12 18:23:38
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061121000051KK02962