证明x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2能被x+y+z整除
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/09/25 11:10:59
证明x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2能被x+y+z整除
证明:∵x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2
=x^4+y^4+z^4+2x^2y^2-2x^2z^2-2y^2z^2-4x^2y^2
=(x^2+y^2-z^2)^2-4x^2y^2
=(x^2+y^2-z^2+2xy)(x^2+y^2-z^2-2xy)
=[(x+y)^2-z^2][(x-y)^2-z^2]
=(x+y+z)(x+y-z)(x-y+z)(x-y-z)
∴(x+y+z)是x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2的一个因式
∴x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2能被x+y+z整除
=x^4+y^4+z^4+2x^2y^2-2x^2z^2-2y^2z^2-4x^2y^2
=(x^2+y^2-z^2)^2-4x^2y^2
=(x^2+y^2-z^2+2xy)(x^2+y^2-z^2-2xy)
=[(x+y)^2-z^2][(x-y)^2-z^2]
=(x+y+z)(x+y-z)(x-y+z)(x-y-z)
∴(x+y+z)是x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2的一个因式
∴x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2能被x+y+z整除
证明:x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2能被(x+y+z)整除
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,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)
(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
求解2x≥x+y+z 4y≥x+y+Z 8z≥x+y+z x、y、z均大于0
(x-2y+z)(x+y-2z)分之(y-x)(z-x) + (x+y-2z)(y+z-2x)分之(z-y)(x-y)
x=y/z=z/3,x+y+z =12,求2x+3y+4z是多少,
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
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(xy-2z)(y+z-2x)+(x-
已知(x+y+z)^2=x^2+y^2+z^2,证明x(y+z)+y(z+x)+z(x+y)=0