若x,y,z均是正整数,试说明(z^2-x^2-y^2)^2-4x^2y^2能被x+y+Z整除
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)
证明:x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2能被(x+y+z)整除
(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x-2z+y)(y+z-2x)+(x
(x-2y+z)(x+y-2z)分之(y-x)(z-x) + (x+y-2z)(y+z-2x)分之(z-y)(x-y)
求解2x≥x+y+z 4y≥x+y+Z 8z≥x+y+z x、y、z均大于0
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(xy-2z)(y+z-2x)+(x-
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+(
已知3x+2y=4+z,2x+2z=6+y,问是否存在x、y、z的正整数值,使得x+y+z
化简:(x+y-z)^3n*(z-x-y)^2n*(x-z+y)^5n(n为正整数)
(x+y+z)^2-(x-y-z)^2