Ta có xy³ + y² + 4xy = 6

⇔ xy³ + y² + 4xy + 4 = 6 + 4

⇔ y²(xy + 1) + 4(xy + 1) = 10

⇔ (xy + 1)(y² + 4) = 10 = 5 . 2 = (– 5) . (– 2)

Vì y² + 4 ≥ 4 với mọi y

Nên \(\left\{ \begin{array}{l}xy + 1 = 2\\{y^2} + 4 = 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}xy = 1\\{y^2} = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}xy = 1\\\left[ \begin{array}{l}y = 1\\y = - 1\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 1\\y = 1\end{array} \right.\\\left\{ \begin{array}{l}x = - 1\\y = - 1\end{array} \right.\end{array} \right.\)

Vậy (x; y) ∈ {(1; 1); (–1; –1)}.