phân tích đa thức thành nhân tử a) x^3+y(1-3x^2)+x(3y^2-1)-y^3 b)a^x+a^2y-7x-7y c)x(x+1)^2+x(x-5)-5(x+1)^2 05/07/2021 Bởi Daisy phân tích đa thức thành nhân tử a) x^3+y(1-3x^2)+x(3y^2-1)-y^3 b)a^x+a^2y-7x-7y c)x(x+1)^2+x(x-5)-5(x+1)^2
$a)x^3+y.(1-3x^2)+x.(3y^2-1)-y^3_{}$ $x^3-3x^2y+3xy^2-y^3+y-x_{}$ $⇔(x-y)^3-(x-y)_{}$ $b)a^2x+a^2y-7x-7y_{}$ $⇔x.(a^2-7)+y.(a^2-7)_{}$ $⇔(a^2-7)(x+y)_{}$ $c)x.(x+1)^2+x(x-5)-5(x+1)^2_{}$ $⇔(x+1)^2.(x-5)+x.(x-5)_{}$ $⇔[ (x+1)(x-5)+x].(x-5).(x+1)_{}$ $⇔(x^2-3x-5)..(x-5)(x+1)_{}$ Bình luận
$a)x^3+y.(1-3x^2)+x.(3y^2-1)-y^3_{}$
$x^3-3x^2y+3xy^2-y^3+y-x_{}$
$⇔(x-y)^3-(x-y)_{}$
$b)a^2x+a^2y-7x-7y_{}$
$⇔x.(a^2-7)+y.(a^2-7)_{}$
$⇔(a^2-7)(x+y)_{}$
$c)x.(x+1)^2+x(x-5)-5(x+1)^2_{}$
$⇔(x+1)^2.(x-5)+x.(x-5)_{}$
$⇔[ (x+1)(x-5)+x].(x-5).(x+1)_{}$
$⇔(x^2-3x-5)..(x-5)(x+1)_{}$
=))
`a)`
`x^3+y(1-3x^2)+x(3y^2-1)-y^3`
`= (x-y)^3-(x-y)`
`= (x-y)(x-y+1)(x-y-1)`