x,y,z为有理数已知x、y、z为有理数,且（y-z)²+（z-x)²+（x-y)²=（z+y-2x）²+（z+x-2y）²+（x+y-2z)².求（yz+1)(zx+1)(xy+1)除以（x²+1)(y²+1)(z²+1)的值

Question

x,y,z为有理数
已知x、y、z为有理数,且（y-z)²+（z-x)²+（x-y)²=（z+y-2x）²+（z+x-2y）²+（x+y-2z)².求（yz+1)(zx+1)(xy+1)除以（x²+1)(y²+1)(z²+1)的值

Answer 1

(x+y-2z)^2+(y+z-2x)^2+(z+x-2y)^2
=((x-z)+(y-z))²+((y-x)+(z-x))²+((z-y)+(x-y))²
=(x-z)²+(y-z)²+2(x-z)(y-z)+(y-x)²+(z-x)²+2(y-x)(z-x)+(z-y)²+(x-y)²+2(z-y)(x-y)
=2(z-x)²+2(y-z)²+2(x-y)²+2(x-z)(y-z)+2(y-x)(z-x)+2(z-y)(x-y)
=(x-y)^2+(y-z)^2+(z-x)^2
所以
-((x-y)^2+(y-z)^2+(z-x)^2)=2(x-z)(y-z)+2(y-x)(z-x)+2(z-y)(x-y)
-2x²-2y²+2xy-2z²+2yz+2zx=2xy-2xz-2zy+2z²+2yz-2xy-2xz+2x²+2xz-2zy-2xy+2y²
-2x²-2y²+2xy-2z²+2yz+2zx=-2xz-2zy+2z²-2xy+2x²+2y²
4x²+4y²+4z²=4xy+4zy+4xz
即x²+y²+z²-xy-zy-xz=0
(1/2)(x²+y²-2xy+x²+z²-2xz+y²+z²-2zy)=0
(1/2)((x-y)²+(x-z)²+(y-z)²)=0
所以x=y=z
[(xy+1)(yz+1)(zx+1)]/ [(x^2+1)(y^2+1)(z^2+1)]
=(x²+1)(y²+1)(z²+1)/[(x^2+1)(y^2+1)(z^2+1)]
=1