解:∵(y﹣z)2+(x﹣y)2+(z﹣x)2=(y+z﹣2x)2+(z+x﹣2y)2+(x+y﹣2z)2. ∴(y﹣z)2﹣(y+z﹣2x)2+(x﹣y)2﹣(x+y﹣2z)2+(z﹣x)2﹣(z+x﹣2y)2=0, ∴(y﹣z+y+z﹣2x)(y﹣z﹣y﹣z+2x)+(x﹣y+x+y﹣2z)(x﹣y﹣x﹣y+2z)+(z﹣x+z+x﹣2y)(z﹣x﹣z﹣x+2y)=0, ∴x2+y2+z2﹣2xy﹣2xz﹣2yz=0, ∴(x﹣y)2+(x﹣z)2+(y﹣z)2=0. ∵x,y,z均为实数, ∴x=y=z. ∴==1. |