试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)

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试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2

试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)

试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
有这样的公式：a^3+b^3+c^2-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) 左边减右边,证明：（x+y-2z)^3+(y+z-2x)^3+(z+x-2y)^3-3（x+y-2z)(y+z-2x)(z+x-2y) =[（x+y-2z)+(y+z-2x)+(z+x-2y)][（x+y-2z)^2+(y+z-2x)^2+(z+x-2y)^2-(x+y-2z)(y+z-2x)- (y+z-2x)(z+x-2y)-(z+x-2y)(x+y-2z)] =0 所以（x+y-2z)^3+(y+z-2x)^3+(z+x-2y)^3=3（x+y-2z)(y+z-2x)(z+x-2y) 回答人：潇湘诗社 ☆國士無雙卍有疑问欢迎追问,

x-z=a,y-z=b,x-y=c 左边=(X+Y-2Z)^3+ (Y+Z-2X)^3 +（Z+X-2Y)^3 =(a+b)^3-(c+a)^3+(c-b)^3 =a^3+3a^2b+3ab^2+b^3-(a^3+3a^2c+3ac^2+c^3)+c^3-3c^2b+3cb^2-b^3 =3a^2b+3ab^2-3a^2c-3ac^2-3c^2b+3cb^2 =3(a^2b+ab^2-a^2c-ac^2-c^2b+cb^2) =3[a^2(b-c)+a(b+c)(b-c)+bc(b-c)] =3(b-c)(a^2+ab+ac+bc) =3(b-c)(a+c)(a+b) =右边.

试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) 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)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)不是+是 * 用行列式的性质证明：y+z z+x x+y x y z x+y y+z z+x =2 z x y z+x x+y y+z y z x 这个怎么证? 证明：x/(y+z)+y/(z+x)+z/(x+y)>=3/2其中 x,y,z>0 试证明(x+y-2z)³+(y+z-2x)³+(z+x-2y)³=3(x+y-2z)(y+z-2x)(z+x-2y) 已知(x+y+z)^2=x^2+y^2+z^2,证明x(y+z)+y(z+x)+z(x+y)=0 定义:x*y= + :x,y,z∈Z,试证明:(x*y) *z=x*(y*z) 不成立定义: x*y= x的（3+a）次方+y的2a次方 : x,y,z∈Z, 试证明: (x*y) *z=x*(y*z) <结合律>不成立不好意思，前面复制漏了证明已知xyz∈R^+, x^2x * y^2y* z^2z≥x^y+x* y^z+x * 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-z)(y-z)/(y+z-2x)(x-2y+z) 计算:(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z) 计算:x^2/(x-y)(x-z)+y^2/(y-x)(y-z)+z^2/(z-x)(z-y) 证明:(y+z-2x)^3+(z+x-2y)^3+(x+y-2z)^3=3(y+z-2x)(z+x-2y)(x+y-2z)注意：是证明不是解答! 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(xy-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)速速回答证明是否恒等式（X-Y)(Y-Z)(Z-X)=XYZ-X^2-Y^2-Z^2 证明1/2x+1/2y+1/2z≥1/(x+y)+1/(x+z)+1/(z+y) (Z-X)-4(X-Y)(Y-Z)=0 证明X-2Y+Z=0 ‘’是平方 x+y+z+2=xyz,x,y,z.为正实数,证明：xyz(x-1)(y-1)(z-1) 化简(x+Y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z)