已知xyz满足z+y+z=xyz 求证:
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已知xyz满足z+y+z=xyz 求证:
x(1-y2)(1-z2)+y(1-x2)(1-z2)+z(1-x2)(1-y2)=4xyz
x(1-y2)(1-z2)+y(1-x2)(1-z2)+z(1-x2)(1-y2)=4xyz
x(1-y²)(1-z²)+y(1-x²)(1-z²)+z(1-x²)(1-y²)
=x(1-y²-z²+y²z²)+y(1-x²-z²+x²z²)+z(1-x²-y²+x²y²)
=x-xy²-xz²+xy²z²+y-yx²-yz²+yx²z²+z-zx²-zy²+zx²y²
=x+y+z-x²y-xy²-x²z-xz²-y²z-yz²+(x+y+z)yz+(x+y+z)xz+(x+y+z)xy
=xyz+xyz+xyz+xyz
=4xyz
=x(1-y²-z²+y²z²)+y(1-x²-z²+x²z²)+z(1-x²-y²+x²y²)
=x-xy²-xz²+xy²z²+y-yx²-yz²+yx²z²+z-zx²-zy²+zx²y²
=x+y+z-x²y-xy²-x²z-xz²-y²z-yz²+(x+y+z)yz+(x+y+z)xz+(x+y+z)xy
=xyz+xyz+xyz+xyz
=4xyz
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