Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\)
\(\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=7k\end{matrix}\right.\)
Ta có: \(xy=40\)
\(\Leftrightarrow2k.5k=40\)
\(\Leftrightarrow10.k^2=40\)
\(\Leftrightarrow k^2=40:10\)
\(\Leftrightarrow k^2=4\)
\(\Leftrightarrow k=2\) hoặc \(k=-2\)
Thay \(k=2\) ta có:
\(\left\{{}\begin{matrix}x=2k=2.2=4\\y=5k=5.2=10\end{matrix}\right.\)
Thay \(k=-2\) ta có:
\(\left\{{}\begin{matrix}x=2k=2.\left(-2\right)=-4\\y=5k=5.\left(-2\right)=-10\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=4;y=10\\x=-4;y=-10\end{matrix}\right.\)