Решите систему уравнений: x^2-3y^2=13; xy=-4.

\[\left\{ \begin{matrix}
x^{2} - 3y^{2} = 13 \\
xy = - 4\ \ \ \ \ \ \ \ \ \ \ \ \\
\end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix}
\left( - \frac{4}{y} \right)^{2} - 3y^{2} = 13 \\
x = \frac{- 4}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\
\end{matrix}\text{\ \ \ \ \ } \right.\ \ \]

\[\left\{ \begin{matrix}
\frac{16}{y^{2}} - 3y^{2} = 13\ \ \ | \cdot y^{2} \\
x = \frac{- 4}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\
\end{matrix} \right.\ \]

\[- 3y^{4} - 13y^{2} + 16 = 0;\ \ \ \ y \neq 0\]

\[Пусть\ \ y^{2} = t:\]

\[- 3t^{2} - 13t + 16 = 0\]

\[\ D = 169 + 192 = 361\]

\[t_{1} = \frac{13 - 19}{- 6} = 1;\ \ \]

\[t_{2} = \frac{13 + 19}{- 6} = - \frac{16}{3}\]

\[y^{2} = 1\ \ \ или\ \ \ \ \]

\[y^{2} = - \frac{16}{3} - не\ удовлетворяет.\]

\[\left\{ \begin{matrix}
y = 1\ \ \ \\
x = - 4 \\
\end{matrix}\text{\ \ \ \ \ } \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix}
y = - 1 \\
x = 4\ \ \ \\
\end{matrix} \right.\ \]

\[Ответ:( - 4;1),(4;\ - 1).\]