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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?
最佳解答:
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...
其他解答:
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|||||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 ...
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