\(\dfrac{x-y}{3}=\dfrac{x+y}{13}\Leftrightarrow13\left(x-y\right)=3\left(x+y\right)\\ \Leftrightarrow13x-13y=3x+3y\\ \Leftrightarrow10x-16y=0\\ \Leftrightarrow2\left(5x-8y\right)=0\Leftrightarrow5x-8y=0\\ \Leftrightarrow5x=8y\Leftrightarrow x=\dfrac{8y}{5}\)
\(\dfrac{xy}{200}=\dfrac{x+y}{13}\Leftrightarrow\dfrac{\dfrac{8y}{5}\cdot y}{200}=\dfrac{\dfrac{8y}{5}+y}{13}\Leftrightarrow\dfrac{2y^2}{250}=\dfrac{\dfrac{13y}{5}}{13}\\ \Leftrightarrow\dfrac{2y^2}{250}=\dfrac{y}{5}\Leftrightarrow10y^2=250y\Leftrightarrow10y^2-250y=0\\ \Leftrightarrow10y\left(y-25\right)=0\Leftrightarrow\left[{}\begin{matrix}y=0\\y=25\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{8\cdot25}{5}=40\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left\{\left(0;0\right);\left(40;25\right)\right\}\)
