Sửa: \(x^2+y^2=369\)
\(x:y=4:5\Rightarrow\dfrac{x}{4}=\dfrac{y}{5}\Rightarrow\dfrac{x^2}{16}=\dfrac{y^2}{25}=\dfrac{x^2+y^2}{16+25}=\dfrac{369}{41}=9\\ \Rightarrow\left\{{}\begin{matrix}x^2=144\\y^2=225\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm12\\y=\pm15\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(12;15\right);\left(-12;-15\right)\)