xy(x-y)-xz(x+z)-yz(2x+y-z) x(y^2-z^2)-y(z^2-y^2)+z(x^2-y^2) x(y+z)^2+y(z+x)^2+z(x+y)^2-4xyz yz(y+z)+xz(z-x)-xy(x+y) giúp em với ạ em hết điểm rồi

xy(x-y)-xz(x+z)-yz(2x+y-z)
x(y^2-z^2)-y(z^2-y^2)+z(x^2-y^2)
x(y+z)^2+y(z+x)^2+z(x+y)^2-4xyz
yz(y+z)+xz(z-x)-xy(x+y) giúp em với ạ em hết điểm rồi

0 bình luận về “xy(x-y)-xz(x+z)-yz(2x+y-z) x(y^2-z^2)-y(z^2-y^2)+z(x^2-y^2) x(y+z)^2+y(z+x)^2+z(x+y)^2-4xyz yz(y+z)+xz(z-x)-xy(x+y) giúp em với ạ em hết điểm rồi”

  1. Sửa đề `a)xy(x-y)-xz(x+z)-yz(2x+y-z)→xy(x-y)-xz(x+z)+yz(2x-y+z)`

    `a)xy(x-y)-xz(x+z)+yz(2x-y+z)`

    `=xy(x-y)-xz(x+z)+yz(x-y+x+z)`

    `=xy(x-y)-xz(x+z)+yz[(x-y)+(x+z)]`

    `=xy(x-y)-xz(x+z)+yz(x-y)+yz(x+z)`

    `=[xy(x-y)+yz(x-y)]-[xz(x+z)-yz(x+z)]`

    `=y(x-y)(x+z)-z(x+z)(x-y)`

    `=(x-y)(x+z)(y-z)`

    `b)x(y^2-z^2)-y(z^2-y^2)+z(x^2-y^2)`

    `=xy^2-xz^2-yz^2+y^3+x^2z-y^2z`

    `=(xy^2+y^3)-(xz^2+yz^2)+(x^2z-y^2z)`

    `=y^2(x+y)-z^2(x+y)+z(x^2-y^2)`

    `=y^2(x+y)-z^2(x+y)+z(x-y)(x+y)`

    `=(x+y)(y^2-z^2+xz-yz)`

    `c)x(y+z)^2+y(z+x)^2+z(x+y)^2-4xyz`

    `=x(y+z)^2+y(z+x)^2-2xyz+z(x+y)^2-2xyz`

    `=x(y+z)^2+[y(z+x)^2-2xyz]+[z(x+y)^2-2xyz]`

    `=x(y+z)^2+y[(z+x)^2-2xz]+z[(x+y)^2-2xy]`

    `=x(y+z)^2+y(z^2+2xz+x^2-2xz)+z(x^2+2xy+y^2-2xy)`

    `=x(y+z)^2+y(z^2+x^2)+z(x^2+y^2)`

    `=x(y+z)^2+yz^2+x^2y+x^2z+y^2z`

    `=x(y+z)^2+(yz^2+y^2z)+(x^2y+x^2z)`

    `=x(y+z)^2+yz(z+y)+x^2(y+z)`

    `=(y+z)(xy+xz+yz+x^2)`

    `=(y+z)[(xy+x^2)+(xz+yz)]`

    `=(y+z)[x(y+x)+z(x+y)]`

    `=(y+z)(x+y)(x+z)`

    `d)yz(y+z)+xz(z-x)-xy(x+y)`

    `=yz(y+z)+xz^2-x^2z-x^2y-xy^2`

    `=yz(y+z)+(xz^2-xy^2)-(x^2z+x^2y)`

    `=yz(y+z)+x(z^2-y^2)-x^2(z+y)`

    `=yz(y+z)+x(z-y)(z+y)-x^2(z+y)`

    `=(y+z)(yz+xz-xy-x^2)`

    `=(y+z)[(yz+xz)-(xy+x^2)]`

    `=(y+z)[z(y+x)-x(y+x)]`

    `=(y+z)(y+x)(z-x)`

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