一道因式分解题x2y-y2z+z2x-x2z+y2x+z2y-2xyz
热心网友
x2y-y2z+z2x-x2z+y2x+z2y-2xyz =(z^2x+y^2x-2xyz)+(x^2y-x^2z)-(y^2z-z^2y)=x(y-z)^2+x^2(y-z)-yz(y-z)=(y-z)(yx-zx+x^2-yz)=(y-z)=(y-z)(x-z)(x+y)
热心网友
x^2y-y^2z+z^2x-x^2z+y^2x+z^2y-2xyz=-z(x^2+2xy+y^2)+z^2(x+y)+xy(x+y)=(x+y)[-z(x+y)+z^2+xy]=(x+y)(z-x)(z-y)
热心网友
x2y-y2z+z2x-x2z+y2x+z2y-2xyz 因为第一项x2y与第五项y2x相同,第三项与第四项也相同,第二项与第六项相同,所以原式=4xy-2xyz=2xy*(2-z)