3x^2 + 3y^2 + 4xy + 2x - 2y + 2=0 
=> 3.x^2 + 2.x.(2y + 1) + 3y^2 - 2y + 2 = 0
=> 3.[x^2 + 2.x.(2y + 1)/3 + (2y + 1)^2 /9] - (2y + 1)^2 /3 + 3y^2 - 2y + 2 = 0
=> 3.(x + (2y + 1)/3)^2 + 5/3.y^2 - 10/3.y + 5/3 = 0
=> 3(x + (2y + 1)/3)^2 + 5/3.(y - 1)^2 = 0
Vì (x + (2y + 1)/3)^2 >= 0 => 3(x + (2y + 1)/3)^2 >= 0
     (y - 1)^2 >= 0 =>5/3.(y  -1)^2 >= 0
=> 3(x + (2y + 1)/3)^2 + 5/3.(y - 1)^2 >= 0
Dấu "=" xảy ra <=> x + (2y + 1)/3 = y - 1 = 0 => x = -1 ; y = 1
=> P = (-2 + 1)^2019 + (-1 + 2)^2020 + (2 - 1)^2021 = 1