初二数学题

证明：x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2能被x+y+z整除。

热心网友

x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2=x^4+2x^2y^2+y^4-2z^2(x^2+y^2)+z^4-4x^2y^2=(x^2+y^2)^2-2z^2(x^2+y^2)+z^4-4x^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)可见x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2含有因式x+y+z,即x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2能被x+y+z整除

热心网友

x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2=x^4-2(y^2+z^2)x^2+(y^4+z^4+2y^2z^2)-4y^2z^2=[x^4-2(y^2+z^2)x^2+(y^2+z^2)^2] - 4y^2z^2=[x^2-(y^2+z^2)]^2-4y^2z^2=(x^2-y^2-z^2+2y^2z^2)(x^2-y^2-z^2-2y^2z^2)=[x^2-(y^2+z^2-2y^2z^2)][x^2-(y^2+z^2+2y^2z^2)]=[x^2-(y-z)^2][x^2-(y+z)^2]=(x+y-z)(x-y+z)(x+y+z)(x-y-z)所以x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2能被x+y+z整除.

热心网友

=x^4+……-2X^2Z^2+2y^2z^2-4y^2z^2=(x^2-y^2-z^2)^2-4y^2z^2=(x^2-y^2-z^2+2yz)(x^2-y^2-z^2-2yz)=[(x^2-(y-z)^2][(x^2-(y+z)^2]再分解一次就可以了

热心网友

你那个小三角是什么符号 啊 除号？