(x+y-z)^2-(x-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)
(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)^2-(x-y+z)^2=?
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为实数 且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+z-2y)^2+(x+y
分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^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-
x/2=y/3=z/5 x+3y-z/x-3y+z
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x-2z+y)(y+z-2x)+(x
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
(x-2y+z)(x+y-2z)分之(y-x)(z-x) + (x+y-2z)(y+z-2x)分之(z-y)(x-y)
已知(x+y+z)^2=x^2+y^2+z^2,证明x(y+z)+y(z+x)+z(x+y)=0