1樓:手機使用者
當x=y時,原式等於0,故原式含有因子x-y,
又因為原式是關於x,y,z的輪換對稱版式,故原權式還含因子y-z,z-x,
又因為原式為x,y,z的五次式,故可設x2(y-z)3+y2(z-x)3+z2(x-y)3
=(x-y)(y-z)(z-x)[a(x2+y2+z2)+b(xy+yz+zx)]
令x=-1,y=0,z=1得2a-b=-1,
令x=0,y=1,z=2得5a+2b=2,
解得a=0,b=1,
所以x2(y-z)3+y2(z-x)3+z2(x-y)3=(x-y)(y-z)(z-x)(xy+yz+zx).
因式分解:x^2(y+z)+y^2(z+x)+z^2(x+y)-(x^3+y^3+z^3)-2xyz 10
2樓:一朵不知所云
^^^^x^2(y+z)+y^2(z+x)+z^2(x+y)-(x^3+y^3+z^3)-2xyz
=x^2(y+z)+y^2*z+y^2*x+z^2*x+z^2*y-x^3-y^3-z^3-2xyz
=x^2(y+z-x)+x(y^2+z^2-2yz)+(y^2*z+z^2*y-y^3-z^3)
=x^2(y+z-x)+x(y-z)^2+[y^2(z-y)-z^2(z-y)]
=x^2(y+z-x)+x(y-z)^2+(y+z)(y-z)(z-y)
=x^2(y+z-x)+x(y-z)^2-(y+z)(y-z)^2
=x^2(y+z-x)+(y-z)^2(x-y-z)
=(x-y-z)(z-y-x)(z-y+x)
因式分解:x2(y-z)+y2(z-x)+z2(x-y)
3樓:匿名使用者
^^x^專2(y-z)+y^屬2(z-x)+z^2(x-y)=x^2y-x^2z+y^2z-xy^2+z^2(x-y)=xy(x-y)-z(x-y)(x+y)+z^2(x-y)=(x-y)(xy-zx+yz+z^2)
因式分解(x+y+z)^3-x^3-y^3-z^3
4樓:小賤
^參考下面
(x+y+z)^3-x^3-y^3-z^3=3yx^2+3xy^2+3xz^2+3yz^2+3zx^2+3zy^2+6xyz
=3xy(x+y)+3z^2(x+y)+3z(x^2+y^2+2xy)
=3xy(x+y)+3z^2(x+y)+3z(x+y)^2=3(x+y)[xy+z^2+z(x+y)]=3(x+y)[x(y+z)+z(y+z)]=3(x+y)(x+z)(y+z)、
、好評,,o(∩_∩)o謝謝
x^2(y+z)+y^2(z+x)+z^2(x+y)-(x^3+y^3+z^3)-2xyz
5樓:西域牛仔王
x^2(y+z)+y^2(z+x)+z^2(x+y)-(x^3+y^3+z^3)-2xyz
= -(x-y+z)(x+y-z)(x-y-z)
2x3x25x2,因式分解
2x 3 2x x 2 3x 2 2x x 1 x 1 x 2 x 1 x 1 2x 2 2x x 2 x 1 2x 2 3x 2 x 1 x 2 2x 1 解 原式 內 2x 容3 2x x 2 3x 2 2x x 1 x 1 x 2 x 1 x 1 2x 2 2x x 2 x 1 2x 2 3x...
3x 2y 2z 3 2x 4y 3z 3 5x 2y 3z 12三元一次方程組
3x 2y 2z 3.2x 4y 3z 3.5x 2y 3z 12.解 得 8x z 9.2 得 4x 7z 3.由 得 z 8x 9 把 代入 得 x 1把x 1代入 得 z 1把x 1 z 1代入 得 y 2即 方程組的解是 x 1 y 2 z 12x 3y 2z 10.3x 2y 2z 1 2...
x3y3z32xyz怎麼分解因式
是x du3 y zhi3 z 3 3xyz嗎?x 3 y dao3 z 3 3xyz x y z x 2 y 2 z 2 xy yz zx 由x 3 y 3 z 3 x y z x 2 y 2 z 2 內 z 容x 2 y 2 x y 2 z 2 y x 2 z 2 x 3 y 3 z 3 3xy...