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+

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】