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更新1:
係米要將佢expand呢＠＠ 但係我expand完 吾知點計＝ ＝
f3 factorization!!!
回答 (4)
2011-10-10 2:23 am
✔ 最佳答案
x+2 和 y-3 可當成一個數 , 不用 expand , 直接用十字乘法 :(x+2)......... 2(y -3)
........... X
(x+2) ...... - 3(y -3)
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- 3(x+2)(y -3) + 2(x+2)(y -3) = - (x+2)(y -3)
原式 = ( (x+2) + 2(y -3) ) ( (x+2) - 3(y -3) )= (x + 2y - 4) (x - 3y + 11)
2011-10-17 2:10 am
原式 = ( (x+2) + 2(y -3) ) ( (x+2) - 3(y -3) )= (x + 2y - 4) (x - 3y + 11)
參考: me
2011-10-11 1:12 am
設a=x+2 b=y-3
(x+2)^2-(x+2)(y-3)-6(y-3)^2
=a^2-ab-6b^2
=(a-3b)(a+2b)
=[(x+2)-3(y-3)][(x+2)+2(y-3)]
=(x+2-3y+3)(x+2+2y-6)
=(x-3y+5)(x+2y-4)
(x+2)^2-(x+2)(y-3)-6(y-3)^2
=a^2-ab-6b^2
=(a-3b)(a+2b)
=[(x+2)-3(y-3)][(x+2)+2(y-3)]
=(x+2-3y+3)(x+2+2y-6)
=(x-3y+5)(x+2y-4)
2011-10-10 9:23 am
或者 e個方法會更易明白
(x+2)^2-(x+2)(y-3)-6(y-3)^2
Let x+2 = a
y-3 = b
a^2 - ab - 6 b^2
= (a + 2b)(a -3b) <---factorize 完記住放番x+2同y-3落去
= [x+2+2(y-3)] [x+2 - 3(y-3)]
= (x+2+2y-6) (x+2-3y+9)
=(x+2y-4) (x-3y+11)
(x+2)^2-(x+2)(y-3)-6(y-3)^2
Let x+2 = a
y-3 = b
a^2 - ab - 6 b^2
= (a + 2b)(a -3b) <---factorize 完記住放番x+2同y-3落去
= [x+2+2(y-3)] [x+2 - 3(y-3)]
= (x+2+2y-6) (x+2-3y+9)
=(x+2y-4) (x-3y+11)
收錄日期: 2021-04-20 11:44:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111009000051KK00818