设x、y、z为实数,且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+z-2y)^2+(x+y-2z)^2,求[(yz+1)(zx+1)(xy+1)]/[(x^2+1)(y^2+1)(z^2+1)]的值.

Question

设x、y、z为实数,且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+z-2y)^2+(x+y-2z)^2,求[(yz+1)(zx+1)(xy+1)]/[(x^2+1)(y^2+1)(z^2+1)]的值.

Answer 1

由于（y-z)²+(x-y)²+(z-x)²=(y+z-2x)²+(x+z-2y)²+(x+y-2z)²,则
(y+z-2x)²+(x+z-2y)²+(x+y-2z)²-（y-z)²-(x-y)²-(z-x)²=0 =》
(y+z-2x+y-z)(y+z-2x-y+z)+(x+z-2y+x-z)(x+z-2y-x+z)+(x+y-2z+y-x)(x+y-2z-y+x)=0 =》
（2y-2x)(2z-2x)+(2x-2y)(2z-2y)+(2y-2z)(2x-2z)=0 =》
4(y-x)(z-x)+4(x-y)(z-y)+4(y-z)(x-z)=0 =》
2(y-x)²+2(z-x)² +2(z-y)²=0 =》
所以y-x=0,z-x=0,z-y=0
则x=y=z
所以原式=(x²+1)³/(x²+1)³
=1