Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\)
Ta có :
\(\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
\(\Rightarrow xy=54\Leftrightarrow2k.3k=54\)
\(\Rightarrow6k^2=54\)
\(\Rightarrow k^2=54:6\)
\(\Rightarrow k^2=9\)
\(\Rightarrow k^2=\left(\pm3\right)^2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(\pm3\right)=\left(\pm6\right)\\y=3.\left(\pm3\right)=\left(\pm9\right)\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(6;9\right),\left(-6;-9\right)\right\}\)