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

Ответ:

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

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

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

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

\[Пусть\ \ y^{2} = t:\]

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

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

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

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

\[y^{2} = 1\ \ \ или\ \ \ \ \]

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

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

\[Ответ:( - 4;1),(4;\ - 1).\]