Đặt \(\dfrac{x}{3}=\dfrac{y}{7}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=7k\end{matrix}\right.\)
Mà \(x.y=84\)
\(\Rightarrow3k.7k=84\)
\(\Rightarrow21k^2=84\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=\pm2\)
+)Nếu \(k=-2\Rightarrow\left\{{}\begin{matrix}x=3.\left(-2\right)=-6\\y=7.\left(-2\right)=-14\end{matrix}\right.\)
+)Nếu \(k=2\Rightarrow\left\{{}\begin{matrix}x=3.2=6\\y=7.2=14\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(-6;-14\right),\left(6;14\right)\right\}\)