Cho $A = {2019^{n + 1}} - {2019^n}$ . Khi đó $A$ chia hết cho số nào dưới đây với mọi $n \in \mathbb{N}$.

$m\left( {x + y + 1} \right)$.

$m\left( {x + y + m} \right)$.

$m\left( {x + y} \right)$.

$m\left( {x + y - 1} \right)$

$4{x^3}{y^2} - 8{x^2}{y^3} = 4{x^2}y\left( {xy - 2{y^2}} \right)$.

$4{x^3}{y^2} - 8{x^2}{y^3} = 4{x^2}{y^2}\left( {x - y} \right)$.

$4{x^3}{y^2} - 8{x^2}{y^3} = 4{x^2}{y^2}\left( {x - 2y} \right)$.

$4{x^3}{y^2} - 8{x^2}{y^3} = 4{x^2}{y^3}\left( {x - 2y} \right)$.

${\left( {x - 2} \right)^2} - {\left( {2 - x} \right)^3} = {\left( {x - 2} \right)^2}\left( {x - 1} \right)$.

${\left( {x - 2} \right)^2} - \left( {2 - x} \right) = \left( {x - 2} \right)\left( {x - 1} \right)$.

${\left( {x - 2} \right)^3} - {\left( {2 - x} \right)^2} = {\left( {x - 2} \right)^2}\left( {3 - x} \right)$.

${\left( {x - 2} \right)^2} + x - 2 = \left( {x - 2} \right)\left( {x - 1} \right)$.

$5x\left( {x - y} \right) - \left( {y - x} \right) = \left( {x - y} \right)\left( {5x + 1} \right)$.

$5x\left( {x - y} \right) - \left( {y - x} \right) = 5x\left( {x - y} \right)$.

$5x\left( {x - y} \right) - \left( {y - x} \right) = \left( {x - y} \right)\left( {5x - 1} \right)$.  

$5x\left( {x - y} \right) - \left( {y - x} \right) = \left( {x + y} \right)\left( {5x - 1} \right)$.

$2a + b$.

$1 + b$.

${a^2} + ab$.

$a + b$.

$x + 2$.

$3\left( {x - 2} \right)$.

${\left( {x - 2} \right)^2}$.

${\left( {x + 2} \right)^2}$.

$x = 3;\,x =  - \dfrac{1}{2}$.

$x =  - 3;\,x =  - \dfrac{1}{2}$.

$x = 3;\,x = \dfrac{1}{2}$.

$x =  - 3;\,x = \dfrac{1}{2}$.