以知x+y-z/z=x-y+z/y=-x+y+z/x且xyz不等于0,则(x+y)(y+x)(z+x)/xyz=

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以知x+y-z/z=x-y+z/y=-x+y+z/x且xyz不等于0,则(x+y)(y+x)(z+x)/xyz=
以知x+y-z/z=x-y+z/y=-x+y+z/x且xyz不等于0,则(x+y)(y+x)(z+x)/xyz=

以知x+y-z/z=x-y+z/y=-x+y+z/x且xyz不等于0,则(x+y)(y+x)(z+x)/xyz=
x+y-z/z=x-y+z/y=-x+y+z/x=k
x+y=z(k+1)
y+z=x(k+1)
x+z=y(k+1)
2x+2y+2z=z(k+1)+x(k+1)+y(k+1)
(k-1)(x+y+z)=0
x+y+z=0
k=1
(x+y)(y+x)(z+x)/xyz
=(-z)(-y)(-x)/xyz
=-1