作业帮已知x、y、z满足x+y+z=xyz,求证：x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)=4xyz
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已知x、y、z满足x+y+z=xyz,求证：x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)=4xyz
已知x、y、z满足x+y+z=xyz,求证：x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)=4xyz
已知x、y、z满足x+y+z=xyz,求证：x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)=4xyz
证明：
x(1-y²)(1-z²)+y(1-x²)(1-z²)+z(1-x²)(1-y²)
=x+y+z-x(y²+z²)+xy²z²-y(x²+z²)+yx²z²-z(x²+y²)+zx²y²(xy²z²,yx²z²,zx²y²中分别提取出xyz,用x+y+z代替,并重新组合各项的顺序)
=xyz+(x+y+z)yz-y²z-yz²+(x+y+z)xz-x²z-xz²+(x+y+z)xy-x²y-xy²
=xyz+xyz+xyz+xyz
=4xyz
因为x+y+z=xyz,所以 左边=x(1-z²-y²+y²z²)+y(1-z²-x²+x²z²)+z(1-y²-x²+x²y²)=(x+y+z)-xz²-xy²+xy²z²-yz²-yx²+yx²z...
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因为x+y+z=xyz,所以 左边=x(1-z²-y²+y²z²)+y(1-z²-x²+x²z²)+z(1-y²-x²+x²y²)=(x+y+z)-xz²-xy²+xy²z²-yz²-yx²+yx²z²-zy²-zx²+zx²y²=xyz-xy(y+x)-xz(z+x)-yz(y+z)+xyz(xy+xz+yz)=xyz-xy(xyz-z)-xz(xyz-y)-yz(xyz-x)+xyz(xy+xz+yz)=xyz+xyz+xyz+xyz=4xyz
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等式左边=
（1-y^2)[x(1-z^2)+z(1-x^2)]+y(1-x^2)(1-z^2)
=(1-y^2)(x+z)(1-zx)+y(1-x^2)(1-z^2)
有x+y+z=xyz 则 x+z=xyz-y
则左边继续=(1-y^2)(xz-1) y(1-zx)+y(1-x^2)(1-z^2)
将里面乘出来=y [1-z^2-x^2+(xz)^2-(...
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等式左边=
（1-y^2)[x(1-z^2)+z(1-x^2)]+y(1-x^2)(1-z^2)
=(1-y^2)(x+z)(1-zx)+y(1-x^2)(1-z^2)
有x+y+z=xyz 则 x+z=xyz-y
则左边继续=(1-y^2)(xz-1) y(1-zx)+y(1-x^2)(1-z^2)
将里面乘出来=y [1-z^2-x^2+(xz)^2-(xz)^2+2zx-1+(yzx)^2-2xzy^2+y^2]
=y [(xyz)(xyz-2y)+y^2-z^2-x^2+2zx]
=y[(x+y+z)(x+z-y)+y^2-z^2-x^2+2zx]
=y[(x+z)^2-y^2+y^2-z^2-x^2+2zx
=4xyz=右边
等式成立
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