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=?

题目

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=?

答案

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.