Ta có
$P = x^3 - x^2y - x^2 - xy^2 + y^3 + y^2 + 6x + 6y + 2020$
$= (x^3 - x^2y) + (y^3 - xy^2) + (y^2 - x^2) + 6x + 6y + 2020$
$= x^2(x-y) -y^2(x - y)+ y^2 - x^2+ 6x + 6y + 2020$
$= x^2 - y^2 - x^2 + y^2 + 6x + 6y +2020$
$= 6(x+y) + 2020$
Vậy $P = 6(x+y) + 2020$
Ta có
$Q = x^2 - xy + y - 2x + 5$
$= x(x-y) + y - 2x + 5$
$= y - x + 5$
$= -(x-y) + 5$
$= -1 + 5 = 4$
Vậy $Q = 4$