z=mx+ny(over)m+n [n]
z=mx+ny/m+n
z(m+n)=mx+ny
zm+zn-ny-mx=0
m(z-x)+n(z-y)=0
n(z-y)=-m(z-x)
n=-m(z-x)/(z-y)
我計得岩唔岩?
主項變換~
2007-12-22 3:20 am
回答 (3)
2007-12-22 5:33 am
✔ 最佳答案
You are right...z=mx+ny/m+n
z(m+n)=mx+ny
zm+zn-ny-mx=0
m(z-x)+n(z-y)=0
n(z-y)=-m(z-x)
n=-m(z-x)/(z-y) or
n=[m(x-z)]/(z-y) or
n=zm-mx/y-z
上二位are also right...
參考: My Maths knowledge
2007-12-22 3:37 am
錯...
z=mx+ny(over)m+n [n]
z(m+n)=mx+ny
zm+zn=mx+ny
zm+zn-zn-mx=mx+ny-mx-zn
zm-mx=ny-zn
zm-mx=n(y-z)
zm-mx
---------=n
y-z
z=mx+ny(over)m+n [n]
z(m+n)=mx+ny
zm+zn=mx+ny
zm+zn-zn-mx=mx+ny-mx-zn
zm-mx=ny-zn
zm-mx=n(y-z)
zm-mx
---------=n
y-z
參考: me
2007-12-22 3:36 am
我吾係好知= =" s o r ...
我計倒嘅係咁...
z=(mx+ny)/(m+n) [n]
z(m+n)=mx+ny
zm+zn=mx+ny
zn-ny=mx-zm
n(z-y)=m(x-z)
n=[m(x-z)]/(z-y)
我個答案同你個答案有d吾同= ="
我計錯吾好怪我... s o r ...
我計倒嘅係咁...
z=(mx+ny)/(m+n) [n]
z(m+n)=mx+ny
zm+zn=mx+ny
zn-ny=mx-zm
n(z-y)=m(x-z)
n=[m(x-z)]/(z-y)
我個答案同你個答案有d吾同= ="
我計錯吾好怪我... s o r ...
參考: 吾知= =你努力d啦~~奸爸爹!!
收錄日期: 2021-04-19 00:13:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071221000051KK02768