If (a^2 - nb^2)(c^2 - nd^2) = X^2 - nY^2.
Find X and Y in terms of a, b, c, d and n.
Identity
2009-02-05 6:34 am
回答 (3)
2009-02-05 8:20 am
✔ 最佳答案
(a^2 - nb^2)(c^2 - nd^2)= [(ac)^2 + (nbd)^2] - n[(ad)^2 + (bc)^2]
= [(ac)^2 + 2nabcd + (nbd)^2] - n[(ad)^2 + 2abcd + (bc)^2]
= [ac + nbd]^2 - n[ad + bc]^2
X = ac + nbd
Y = ad + bc
參考: ME
2009-02-05 10:20 pm
這樣都得:
(a^2 - nb^2)(c^2 - nd^2)
= [(ac)^2 + (nbd)^2] - n[(ad)^2 + (bc)^2]
= [(ac)^2 - 2nabcd + (nbd)^2] - n[(ad)^2 -2abcd + (bc)^2]
= [ac - nbd]^2 - n[ad - bc]^2
So X=±(ac-nbd)
Y=±(ad-bc)
這題可以有很多答案。
(a^2 - nb^2)(c^2 - nd^2)
= [(ac)^2 + (nbd)^2] - n[(ad)^2 + (bc)^2]
= [(ac)^2 - 2nabcd + (nbd)^2] - n[(ad)^2 -2abcd + (bc)^2]
= [ac - nbd]^2 - n[ad - bc]^2
So X=±(ac-nbd)
Y=±(ad-bc)
這題可以有很多答案。
2009-02-05 8:52 am
Expand to both sides
(a^2 - nb^2)(c^2 - nd^2) = X^2 - nY^2.
[(ac)^2+(nbd)^2]-n[(ad)^2+(bc)^2]=X^2 - nY^2
So
X=√[(ac)^2+(nbd)^2]
Y=√[(ad)^2+(bc)^2]
哈﹐我以後要留心些了
(a^2 - nb^2)(c^2 - nd^2) = X^2 - nY^2.
[(ac)^2+(nbd)^2]-n[(ad)^2+(bc)^2]=X^2 - nY^2
So
X=√[(ac)^2+(nbd)^2]
Y=√[(ad)^2+(bc)^2]
哈﹐我以後要留心些了
收錄日期: 2021-04-21 22:02:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090204000051KK02600