$\dfrac{x}{2}=\dfrac{2y}{3}$
$→x=\dfrac{4y}{3}$
Thế $x=\dfrac{4y}{3}$ vào biểu thức $xy$
$\dfrac{4y}{3}.y=27$
$↔4y²=81$
$↔y²=\dfrac{81}{4}$
\(\leftrightarrow\left[ \begin{array}{l}y=\dfrac{9}{2}\\y=-\dfrac{9}{2}\end{array} \right.\)
\(\to \left[ \begin{array}{l}x=6\\x=-6\end{array} \right.\)
Vậy $(x;y)=\bigg\{\bigg(6;\dfrac{9}{2}\bigg);\bigg(-6;-\dfrac{9}{2}\bigg)\bigg\}$
