已知xyz≠0,x≠y,如果(x^2-yz)/[(x(1-yz)]=(y^2-xz)/[y(1-xz)]成立,求证：x+y+z=1/x+1/y+1/z.

已知xyz≠0,x≠y,如果(x^2-yz)/[(x(1-yz)]=(y^2-xz)/[y(1-xz)]成立,求证：x+y+z=1/x+1/y+1/z.
已知xyz≠0,x≠y,如果(x^2-yz)/[(x(1-yz)]=(y^2-xz)/[y(1-xz)]成立,求证：x+y+z=1/x+1/y+1/z.

已知xyz≠0,x≠y,如果(x^2-yz)/[(x(1-yz)]=(y^2-xz)/[y(1-xz)]成立,求证：x+y+z=1/x+1/y+1/z.
证明：(x-(yz/x))/(1-yz)=(y-(xz/y))/(1-xz),
十字相乘得：(x-(yz/x))×(1-xz)=（y-(xz/y))×(1-yz),
化简：x-(yz/x)-x²z+yz²=y-(xz/y)-y²z+xz²,
移项：y-x+yz/x-xz/y+x²z-yz²-y²z+xz²=0,
合并同类项（y-x）+z(y/x-x/y)+z(x²-yz-y²+xz)=0,
(y-x)+z(y-x)(y+x)/(xy)+z{(x+y)(x-y)+z(x-y)}=0,
(y-x)(1+z/x+z/y)+z(x-y)(x+y+z)=0,
(y-x){1+z/x+z/y-z(x+y+z)}=0,
因为x≠y,所以1+z/x+z/y-z(x+y+z)=0,即z(x+y+z)=1+z/x+z/y
同÷z,得：x+y+z=1/x+1/y+1/z【证明完毕】

(x^2-yz)/x(1-yz)=(y^2-xz)/y(1-xz)
y(x^2-yz)(1-xz)=x(y^2-xz)(1-yz)
y(x^2-x^3z-yz+xyz^2)=x(y^2-y^3z-xz+xyz^2)
xy(x-y)-xyz(x+y)(x-y)-z(x+y)(y-x)+xyz^2(y-x)=0
两边同除以(x-y)
xy-xyz(x+y)+z(...

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(x^2-yz)/x(1-yz)=(y^2-xz)/y(1-xz)
y(x^2-yz)(1-xz)=x(y^2-xz)(1-yz)
y(x^2-x^3z-yz+xyz^2)=x(y^2-y^3z-xz+xyz^2)
xy(x-y)-xyz(x+y)(x-y)-z(x+y)(y-x)+xyz^2(y-x)=0
两边同除以(x-y)
xy-xyz(x+y)+z(x+y)-xyz^2=0
两边同除以xyz
1/z-x-y+1/y+1/x-z=0
x+y+z=1/x+1/y+1/z
得证。

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