若z分之x+y+z=y分之x-y+z=x分之-x+y+z,求xyz分之(x+y)(y+z)(z+x)
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若z分之x+y+z=y分之x-y+z=x分之-x+y+z,求xyz分之(x+y)(y+z)(z+x)
设(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x =k 则
(1)x+y-z=kz
(2)x-y+z=ky
(3)-x+y+z=kx
(1)+(2)+(3)得x+y+z=k(x+y+z)
∴k=1时,或x+y+z=0
当k=1时,即(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x
∴x+y=xz,x+z=2y,y+z=2x
∴原式=(2z*2x*2y)/xyz=8xyz/xyz=8
当x+y+z=0时,则x+y=-z,y+z=-x,z+x=-y
∴原式=-xyz=xyz=-1
∴原式=8或-1
(1)x+y-z=kz
(2)x-y+z=ky
(3)-x+y+z=kx
(1)+(2)+(3)得x+y+z=k(x+y+z)
∴k=1时,或x+y+z=0
当k=1时,即(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x
∴x+y=xz,x+z=2y,y+z=2x
∴原式=(2z*2x*2y)/xyz=8xyz/xyz=8
当x+y+z=0时,则x+y=-z,y+z=-x,z+x=-y
∴原式=-xyz=xyz=-1
∴原式=8或-1
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