中二因式分解

2007-02-05 2:32 am
(h-k)^2-2(h-k)(h+k)+(h+k)^2

(2c+3d)^2-6(2c+3d)(c-2d)+9(c-2d)^2

回答 (2)

2007-02-05 6:49 am
✔ 最佳答案
(h-k)^2-2(h-k)(h+k)+(h+k)^2
=[(h-k)-(h+k)]^2
=(-2k)^2

↑(x^2-2xy+y^2) x係(h-k)
=(x-y)^2 y係(h+k)
---------------------------------------------------
(2c+3d)^2-6(2c+3d)(c-2d)+9(c-2d)^2
=[(2c+3d)-3(c-2d)]^2
=(-c+9d)^2

↑x^2-6xy+9y^2 x係(2c+3d)
=(x-3y)^2 y係(c-2d)
參考: 自己.....信我...我maths攞193/200分
2007-02-05 3:43 am
(h-k)^2-2(h-k)(h+k)+(h+k)^2
=h^2-2hk+k^2-2h^2-2k^2+h^2+2hk+k^2
=h^2+h^2-2hk+2hk+k^2+k^2-2h^2-2k^2
=2h^2+2k^2-2h^2-2k^2
=0


(2c+3d)^2-6(2c+3d)(c-2d)+9(c-2d)^2
=4c^2+12cd+9d^2-12c^2-6cd-36d^2+9c^2-36cd+9d^2
=4c^2-12c^2+9c^2+12cd-6cd-36cd+9d^2-36d^2+9d^2
=c^2-30cd-36d^2


收錄日期: 2021-04-24 08:29:34
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