Phân tích đa thức thành nhân tử : x^3+y^3+z^3- 3xyz 09/08/2021 Bởi Kinsley Phân tích đa thức thành nhân tử : x^3+y^3+z^3- 3xyz
x^3+y^3+z^3- 3xyz =(x+y)³-3x²y-3xy²+z³-3xyz =[(x+y)³+z³]-3xy(x+y+z) =(x+y+z)[(x+y)²-z(x+y)+z²]-3xy(x+y+z) =(x+y+z)(x²+2xy+y²-xz-yz+z²-3xy) =(x+y+z)(x²+y²+z²-xy-yz-xz) Bình luận
$@Vân$ $x^3+y^3+z^3-3xyz_{}$ $=(x+y)^3-3x^2y-3xy^2+z^3-3xyz_{}$ $=[ (x+y)^3+z^3]-3xy.(x+y+z)_{}$ $=(x+y+z).[ (x+y)^2-z.(x+y)+z^2]-3xy.(x+y+z)_{}$ $=(x+y+z).(x^2+2xy+y^2-xz-yz+z^2-3xy)_{}$ $=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)_{}$ Bình luận
x^3+y^3+z^3- 3xyz
=(x+y)³-3x²y-3xy²+z³-3xyz
=[(x+y)³+z³]-3xy(x+y+z)
=(x+y+z)[(x+y)²-z(x+y)+z²]-3xy(x+y+z)
=(x+y+z)(x²+2xy+y²-xz-yz+z²-3xy)
=(x+y+z)(x²+y²+z²-xy-yz-xz)
$@Vân$
$x^3+y^3+z^3-3xyz_{}$
$=(x+y)^3-3x^2y-3xy^2+z^3-3xyz_{}$
$=[ (x+y)^3+z^3]-3xy.(x+y+z)_{}$
$=(x+y+z).[ (x+y)^2-z.(x+y)+z^2]-3xy.(x+y+z)_{}$
$=(x+y+z).(x^2+2xy+y^2-xz-yz+z^2-3xy)_{}$
$=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)_{}$