已知(x+y-xy)/(x+y+2xy)=(y+z-2yz)/(y+z+3yz)=(z+x-3zx)/(z+x+4zx)
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已知(x+y-xy)/(x+y+2xy)=(y+z-2yz)/(y+z+3yz)=(z+x-3zx)/(z+x+4zx),且2/x=3/y-1/z,则xyz=?
![已知(x+y-xy)/(x+y+2xy)=(y+z-2yz)/(y+z+3yz)=(z+x-3zx)/(z+x+4zx)](/uploads/image/z/18696801-57-1.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%88x%2By-xy%EF%BC%89%2F%28x%2By%2B2xy%29%3D%28y%2Bz-2yz%29%2F%28y%2Bz%2B3yz%29%3D%28z%2Bx-3zx%29%2F%28z%2Bx%2B4zx%29)
令(x+y-xy)/(x+y+2xy)=(y+z-2yz)/(y+z+3yz)=(z+x-3zx)/(z+x+4zx)=k,于是
k=(x+y-xy)/(x+y+2xy)=(1/x+1/y-1)/(1/x+1/y+2),将1/x+1/y作为一个整体解出1/x+1/y=(1+2k)/(1-k).类似可得1/y+1/z=(2+3k)/(1-k),1/z+1/x=(3+4k)/(1-k),于是可得1/x=(1+3k/2)/(1-k),1/y=(k/2)/(1-k),1/z=(2+5k/2)/(1-k),代入2/x=3/y-1/z可解出k=-1,于是可知x=y=z=-4,xyz=-64.
k=(x+y-xy)/(x+y+2xy)=(1/x+1/y-1)/(1/x+1/y+2),将1/x+1/y作为一个整体解出1/x+1/y=(1+2k)/(1-k).类似可得1/y+1/z=(2+3k)/(1-k),1/z+1/x=(3+4k)/(1-k),于是可得1/x=(1+3k/2)/(1-k),1/y=(k/2)/(1-k),1/z=(2+5k/2)/(1-k),代入2/x=3/y-1/z可解出k=-1,于是可知x=y=z=-4,xyz=-64.
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