http://i1233.photobucket.com/albums/ff400/rein1993/672A547D540D-1_zpsfb4a3a5c.jpg
這題我知道它要用哪條公式去拆
但想問(a)如何得出(x+2y)(x^2-2xy+4y^2)這個答案?
主要是不知如何才會拆出那些"2"跟"4"
另外(b)也請解一下
謝謝
數學溫習-因式分解一問
2013-04-02 1:01 am
回答 (3)
2013-04-02 3:55 am
✔ 最佳答案
2.(a)x^3 + 8y^3
= x^3 + (2y)^3
= (x + 2y)[x^2 - (x)(2y) + (2y)^2]
= (x + 2y)(x^2 - 2xy + 4y^2)
2.(b)
x^3 + 8y^3 + xz + 2yz
= (x^3 + 8y^3) + (xz + 2yz)
= (x + 2y)(x^2 - 2xy + 4y^2) + z(x + 2y)
= (x + 2y)(x^2 - 2xy + 4y^2 + z)
2013-04-03 12:03 am
恆等式(1)
x^3 +/- y^3
=(x +/- y)(x^2 -/+ xy + y^2)
所以:
(1) x^3 + 8y^3
= (x)^3 + (2y)^3
= (x+2y)(x^2 - 2xy + 4y^2)
==> 恆等式(1)
(2) x^3 + 8y^3 + xz + 2yz
*=(x+2y)(x^2 - 2xy + 4y^2)+z(x+2y)
**=(x+2y)(x^2-2xy+4y^2+x+2y)
*==> 用part a的答案
+抽common factor"z"
**==> 抽common factor (x+2y)
2013-04-02 16:08:01 補充：
更正:
(2)
x^3 + 8y^3 + xz + 2yz
*=(x+2y)(x^2 - 2xy + 4y^2)+z(x+2y)
**=(x+2y)(x^2-2xy+4y^2+z)
不好意思
x^3 +/- y^3
=(x +/- y)(x^2 -/+ xy + y^2)
所以:
(1) x^3 + 8y^3
= (x)^3 + (2y)^3
= (x+2y)(x^2 - 2xy + 4y^2)
==> 恆等式(1)
(2) x^3 + 8y^3 + xz + 2yz
*=(x+2y)(x^2 - 2xy + 4y^2)+z(x+2y)
**=(x+2y)(x^2-2xy+4y^2+x+2y)
*==> 用part a的答案
+抽common factor"z"
**==> 抽common factor (x+2y)
2013-04-02 16:08:01 補充：
更正:
(2)
x^3 + 8y^3 + xz + 2yz
*=(x+2y)(x^2 - 2xy + 4y^2)+z(x+2y)
**=(x+2y)(x^2-2xy+4y^2+z)
不好意思
參考: me
2013-04-02 2:06 am
Formula: a³ ± b³ = (a ± b)(a² ∓ ab + b²)
收錄日期: 2021-04-13 19:23:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130401000051KK00213