Question

求助，一道初二奥数题, 急！

已知(y-z)2+(z-x)2+(x-y)2=(x+y-2z)2+(y+z-2x)2+(z+x-2y)2,
求[(yz+1)(zx+1)(xy+1)]/[(x2+1)(y2+1)(z2+1)] 的值

需要运算过程！

Answer 1

(y+z-2x)^2+(z+x-2y)^2+(x+y-2z)^2
=[(y-x)+(z-x)]^2+[(z-y)+(x-y)]^2+[(x-z)+(y-z)]^2
=2(x-y)^2+2(y-z)^2+2(z-x)^2+2(x-y)(z-y)+2(y-x)(z-x)+2(x-z)(y-z)
=(y-z)^2+(z-x)^2+(x-y)^2
所以
(y-z)^2+(z-x)^2+(x-y)^2+2(x-y)(z-y)+2(y-x)(z-x)+2(x-z)(y-z)=0
展开得
4(x^2+y^2+z^2)-3(xy+xz+yz)=0

Answer 2

已知(y-z)2+(z-x)2+(x-y)2=(x+y-2z)2+(y+z-2x)2+(z+x-2y)2,
取特值x=y=z=1成立
[(yz+1)(zx+1)(xy+1)]/[(x2+1)(y2+1)(z2+1)] =8/27

Answer 3

讨论xyz是否相等，0或8/27

Answer 4

1

Answer 5

他几等于0