Phân tích các đa thức sau thành nhân tử: a) (x + y + z) ^3 – x^3 – y^3 – z^3. b) x^4 + 2010x^2 + 2009x + 2010. 09/11/2021 Bởi Alice Phân tích các đa thức sau thành nhân tử: a) (x + y + z) ^3 – x^3 – y^3 – z^3. b) x^4 + 2010x^2 + 2009x + 2010.
Bài làm : a, `(x+y+z)^3−x^3−y^3−z^3` `=x^3+y^3+z^3+3(x+y)(y+z)(z+x)−x^3−y^3−z^3` `=3(x+y)(y+z)(z+x)` b, `x^4-x+2010x²+2009x+x+2010` `=(x^4-x)+(2010x² +2010x+2010)` `=x(x^3-1)+2010(x²+x+1)` `=(x²+x+1)[x(x-1)+2010]` `=(x²+x+1)(x²-x+2010)` Bình luận
`a)` `(x + y + z) ^3 – x^3 – y^3 – z^3` `=x^3+y^3+z^3+3x^2y+3xy^2+3x^2z+3xz^2+3y^2z+3yz^2+6xyz` `=3(xyz+x^2y+x^2z+xz^2+xy^2+y^2z+xyz+yz^2)` `=3[xy(x+z)+xz(x+z)+y^2(x+z)+yz(x+z)]` `=3(x+z)(xy+xz+y^2+yz)` `=3(x+z)[x(y+z)+y(y+z)]` `=3(x+z)(y+z)(x+y)` `b)` `x^4 + 2010x^2 + 2009x + 2010` `=(x^4-x)+(2010x^2+2010x+2010)` `=x(x^3-1)+2010(x^2+x+1)` `=x(x-1)(x^2+x+1)+2010(x^2+x+1)` `=(x^2+x+1)(x^2-x+2010)` Bình luận
Bài làm :
a, `(x+y+z)^3−x^3−y^3−z^3`
`=x^3+y^3+z^3+3(x+y)(y+z)(z+x)−x^3−y^3−z^3`
`=3(x+y)(y+z)(z+x)`
b, `x^4-x+2010x²+2009x+x+2010`
`=(x^4-x)+(2010x² +2010x+2010)`
`=x(x^3-1)+2010(x²+x+1)`
`=(x²+x+1)[x(x-1)+2010]`
`=(x²+x+1)(x²-x+2010)`
`a)` `(x + y + z) ^3 – x^3 – y^3 – z^3`
`=x^3+y^3+z^3+3x^2y+3xy^2+3x^2z+3xz^2+3y^2z+3yz^2+6xyz`
`=3(xyz+x^2y+x^2z+xz^2+xy^2+y^2z+xyz+yz^2)`
`=3[xy(x+z)+xz(x+z)+y^2(x+z)+yz(x+z)]`
`=3(x+z)(xy+xz+y^2+yz)`
`=3(x+z)[x(y+z)+y(y+z)]`
`=3(x+z)(y+z)(x+y)`
`b)` `x^4 + 2010x^2 + 2009x + 2010`
`=(x^4-x)+(2010x^2+2010x+2010)`
`=x(x^3-1)+2010(x^2+x+1)`
`=x(x-1)(x^2+x+1)+2010(x^2+x+1)`
`=(x^2+x+1)(x^2-x+2010)`