$\begin{array}{l} 3y\left( {y + 2x} \right) + 2\left( {y + 2x} \right) = 65\\ \Leftrightarrow \left( {y + 2x} \right)\left( {3y + 2} \right) = 65 \end{array}$
Vì $x,y$ nguyên dương nên ta suy ra $y+2x\ge 0$, $3y+2\ge 2$ nên ta có các trường hợp:
$\begin{array}{l} \left\{ \begin{array}{l} 3y + 2 = 5\\ y + 2x = 13 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} y = 1\\ x = 6 \end{array} \right.\\ \left\{ \begin{array}{l} 3y + 2 = 13\\ y + 2x = 5 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} y = \dfrac{{11}}{3}(L)\\ x = \dfrac{2}{3} \end{array} \right.\\ \left\{ \begin{array}{l} 3y + 2 = 65\\ y + 2x = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} y = 21\\ x = - 10 \end{array} \right.(L)\\ \left\{ \begin{array}{l} 3y + 2 = 1\\ y + 2x = 65 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} y = - \dfrac{1}{3}\\ x = \dfrac{{98}}{3} \end{array} \right.\left( L \right)\\ \Rightarrow \left( {x;y} \right) = \left( {6;1} \right) \end{array}$
