Phân tích đa thức thành nhân tử

N

nguyen_van_ba

Phân tích đa thức thành nhân tử
[tex]1, (x^2+y^2)^3+(z^2-x^2)^3-(y^2+z^2)^3[/tex]
[tex]2, x^3+y^3+z^3-3xyz[/tex]
[tex]3, (a+b+c)^3-a^3-b^3-c^3[/tex]
[TEX]1, (x^2+y^2)^3+(z^2-x^2)^3-(y^2+z^2)^3[/TEX]
[TEX]=(x^2+y^2+z^2-x^2)^{3}-3(x^2+y^2+z^2-x^2)(x^2+y^2)(z^2-x^2)-(y^2+z^2)^3[/TEX]
[TEX]=-3(y^2+z^2)(x^2+y^2)(z^2-x^2)[/TEX]
 
Last edited by a moderator:
A

anh_bo_doi_cu_ho

làm nốt bài 2


$x^3+y^3+z^3-3xyz$
$=(x+y+z)^3-3x^2.y-3x.y^2-3y^2.z-3y.z^2…$
$=(x+y+z)^3-3xy(x+y+z)-3yz(x+y+z)-3xz(x…$
$=(x+y+z)(<x+y+z>^2-3xy-3yz-3xz)$
$=(x+y+z)(x^2+y^2+z^2+2xy+2yz+2xz-3xy-3…$
$=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$
 
Last edited by a moderator:
N

nguyen_van_ba

[TEX]2, x^3+y^3+z^3-3xyz[/TEX]
[TEX]=(x+y)^{3}-3xy(x+y)-3xyz+z^3[/TEX]
[TEX]=(x+y)^{3}+z^3-3xy(x+y+z)[/TEX]
[TEX]=(x+y+z)^{3}-3(x+y)z(x+y+z)-3xy(x+y+z)[/TEX]
[TEX]=(x+y+z)((x+y+z)^{2}-3(xy+yz+zx))[/TEX]
 
Top Bottom