phân tích thành nhân tử: a)(x+y+z)^3-x^3-y^3-z^3 b)x^4+2010 .x^2+2009x+2010 20/08/2021 Bởi Adalyn phân tích thành nhân tử: a)(x+y+z)^3-x^3-y^3-z^3 b)x^4+2010 .x^2+2009x+2010
`a)(x+y+z)^3-x^3 – y^3 – z^3` `=[(x+y+z)^3-x^3] -(y^3+z^3)` `=(x+y+z-x)[(x+y+z)^2 + (x+y+z).x + x^2]-(y+z)(y^2-yz+z^2)` `=(y+z){[(x+y)+z]^2 + x^2 + xy +xz+x^2} – (y+z)(y^2-yz+z^2)` `=(y+z){[(x+y)+z]^2 + x^2 + xy +xz+x^2-(y^2-yz+z^2)} ` `= ( y+z)[(x+y)^2 + 2z(x+y) + z^2 + x^2 + xy +xz+x^2-y^2+yz-z^2]` `=(y+z)(x^2+2xy+y^2+2xz+2zy + z^2 + x^2 + xy +xz+x^2-y^2+yz-z^2)` `=(y+z)(3x^2+3xy+3xz+3zy)` `= 3(y+z)(x^2+xy+xz+zy)` `=3(y+z)[x(x+y)+z(x+y)]` `=3(y+z)(x+y)(x+z).` `b)x^4+2010 .x^2+2009x+2010` `= x^4 – x + 2010x^2 + 2010x + 2010` `= (x^4-x)+2010.(x^2+x+1)` `=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(x-1)+2010]` `=(x^2+x+1)(x^2-x+2020).` Bình luận
`a)(x+y+z)^3-x^3 – y^3 – z^3`
`=[(x+y+z)^3-x^3] -(y^3+z^3)`
`=(x+y+z-x)[(x+y+z)^2 + (x+y+z).x + x^2]-(y+z)(y^2-yz+z^2)`
`=(y+z){[(x+y)+z]^2 + x^2 + xy +xz+x^2} – (y+z)(y^2-yz+z^2)`
`=(y+z){[(x+y)+z]^2 + x^2 + xy +xz+x^2-(y^2-yz+z^2)} `
`= ( y+z)[(x+y)^2 + 2z(x+y) + z^2 + x^2 + xy +xz+x^2-y^2+yz-z^2]`
`=(y+z)(x^2+2xy+y^2+2xz+2zy + z^2 + x^2 + xy +xz+x^2-y^2+yz-z^2)`
`=(y+z)(3x^2+3xy+3xz+3zy)`
`= 3(y+z)(x^2+xy+xz+zy)`
`=3(y+z)[x(x+y)+z(x+y)]`
`=3(y+z)(x+y)(x+z).`
`b)x^4+2010 .x^2+2009x+2010`
`= x^4 – x + 2010x^2 + 2010x + 2010`
`= (x^4-x)+2010.(x^2+x+1)`
`=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(x-1)+2010]`
`=(x^2+x+1)(x^2-x+2020).`