已知x^2-y^2-z^2=0,A是关于x,y,z的一次多项式,且x^3-y^3-z^3=(x-y)(x-z)A,则A的表达式是()
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/25 00:04:13
已知x^2-y^2-z^2=0,A是关于x,y,z的一次多项式,且x^3-y^3-z^3=(x-y)(x-z)A,则A的表达式是()
已知x^2-y^2-z^2=0,A是关于x,y,z的一次多项式,且x^3-y^3-z^3=(x-y)(x-z)A,则A的表达式是()
已知x^2-y^2-z^2=0,A是关于x,y,z的一次多项式,且x^3-y^3-z^3=(x-y)(x-z)A,则A的表达式是()
x^2-y^2-z^2=0
(x^2-y^2-z^2)(x+y+z)=0
x^3-y^3-z^3+x^2*y+x^2*z-y^2*x-y^2*z-z^2*x-z^2*y=0
x^3-y^3-z^3=-x^2*y-x^2*z+y^2*x+y^2*z+z^2*x+z^2*y
-x^2*y-x^2*z+y^2*x+y^2*z+z^2*x+z^2*y=(x-y)(x-z)A
可解
因为x^2-y^2-z^2=0
所以z^3=z(x^2-y^2) y^2=x^2-z^2
因为X^3-Y^3-Z^3=(X-Y)(X-Z)A
所以x^3-y^3-z(x^2-y^2)=(x-y)(x-z)A
则(x-y)(x^2+xy+y^2)-(x-y)(zx+zy)=(X-Y)(x-Z)A
所以(x-y)[x^2+(y-z)x+y^2-yz]=...
全部展开
因为x^2-y^2-z^2=0
所以z^3=z(x^2-y^2) y^2=x^2-z^2
因为X^3-Y^3-Z^3=(X-Y)(X-Z)A
所以x^3-y^3-z(x^2-y^2)=(x-y)(x-z)A
则(x-y)(x^2+xy+y^2)-(x-y)(zx+zy)=(X-Y)(x-Z)A
所以(x-y)[x^2+(y-z)x+y^2-yz]=(X-Y)(x-Z)A
(x-y)[x^2+(y-z)x+x^2-z^2-yz]=(X-Y)(x-Z)A
(x-y)(x-z)(2x+y+z)=(x-y)(x-z)A
所以A=2x+y+z
收起