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+

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