设x,y,z是实数,且(x-y)^2+(y-z)^2+(z-x)^2=(x+y-2z)^2+(y+z-2x)^2+(z+

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/07/07 08:02:18

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

(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

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 x,y,z为实数且(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y- 已知 x,y,z都是正实数,且 x+y+z=xyz 证明 (y+x)/z+(y+z)/x+(z+x)/y≥2(1/x+1 (y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y) 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(x-2z+y)(y+z-2x)+(x 试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) 设x、y、z为整数,证明:x^4*(y-z)+y^4*(z-x)+z^4*(x-y)/(y+z)^2+(z+x)^2+( (x-2y+z)(x+y-2z)分之（y-x)(z-x) + (x+y-2z)(y+z-2x)分之(z-y)(x-y) 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(xy-2z)(y+z-2x)+(x- (x+y-z)^2-(x-y+z)^2=? 若实数x、y、z满足2|x−y|+2y+z+z