中二主項變換

2014-06-11 3:50 am
1.) x= 2 - 1/1+yz (y)


2.) xz= 1 - 1/1-y (y)

回答 (2)

2014-06-12 12:55 am
✔ 最佳答案
x = 2 - 1/ (1+yz)
x(1+yz) = 2(1 +yz) - 1 .... [把所以數x(1+yz), - 1/(1+yz) x (1+yz) 會cancel out = -1]
x + xyz = 2 + 2yz - 1 .... [拆括號]
xyz - 2yz = 2 - 1 - x .... [將同y聯繫既放喺左邊, 其他轉去右邊]
y(xz-2z) = 1 - x .... [抽出y, common factor]
y = (1-x)/(xz-2z) .... [將括號搬去右邊做分母]
y = (1-x)/(z(x-2)) .... [因為分母2個數都有z, 所以要抽出z]

xz= 1 - 1/1-y
xz(1-y) = 1(1-y) - 1 .... [把所以數x(1-y), - 1/(1-y) x (1-y) 會cancel out = -1]
xz - xyz = 1 - y - 1
-xyz + y = -xz
-xyz -(y)(-1) = -xz
-y (xz - 1)= -xz
y(xz - 1) = xz .... [因為2邊都係負數, 所以可以刪除左個-號]
y = xz/(xz-1)
參考: me
2014-06-12 3:07 am
由於小學豬很熱衷回答,所以我把我的答案移到意見區吧。

回答者:知足常樂 ( 知識長 )
擅長領域: 數學 | 數學
回答時間:2014-06-10 20:01:46

Please read:

https://s.yimg.com/rk/HA00430218/o/2085348751.png

2014-06-11 19:09:13 補充:
Re: 發問者 Man fu,我就是上一次寄電郵給你的那個人了:

https://hk.knowledge.yahoo.com/question/question?qid=7014020500094

看意見欄

http://postimg.org/image/k39nyc6tz/


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