Решите систему уравнений: 4x^2-3xy-y^2=14; 2x^2+xy-3y^2=12.

Ответ:

\[\left\{ \begin{matrix} 4x^{2} - 3xy - y^{2} = 14 \\ 2x^{2} + xy - 3y^{2} = 12 \\ \end{matrix}\text{\ \ \ \ } \right.\ \]

\[Пусть\ \ y = tx:\]

\[\left\{ \begin{matrix} 4x^{2} - 3x^{2}t - t^{2}x^{2} = 14 \\ 2x^{2} + tx^{2} - 3t^{2}x^{2} = 12\ \ \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x^{2}\left( 4 - 3t - t^{2} \right) = 14 \\ x^{2}\left( 2 + t - 3t^{2} \right) = 12 \\ \end{matrix} \right.\ \]

\[\frac{4 - 3t - t^{2}}{2 + t - 3t^{2}} = \frac{7}{6}\]

\[24 - 18t - 6t^{2} = 17 + 7t - 21t^{2}\]

\[15t^{2} - 25t + 10 = 0\ \ \ \ \ |\ :5\]

\[3t^{2} - 5t + 2 = 0\]

\[D = 25 - 24 = 1\]

\[t = \frac{5 + 1}{6} = 1,\ \ \ \ \ \ \ \ t = \frac{5 - 1}{6} = \frac{2}{3}\]

\[1.\ \ \left\{ \begin{matrix} y = x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4x^{2} - 3x^{2} - x^{2} = 14 \\ \end{matrix} \right.\ \]

\[0 \cdot x^{2} = 14\]

\[нет\ корней.\]

\[2)\ \left\{ \begin{matrix} y = \frac{2}{3}\text{x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 4x^{2} - \frac{3 \cdot 2}{3}x^{2} - \frac{4}{9}x^{2} = 14 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 36x^{2} - 18x^{2} - 4x^{2} = 126 \\ y = \frac{2}{3}\text{x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix}\ \right.\ \]

\[\text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} 14x^{2} = 126 \\ y = \frac{2}{3}\text{x\ \ \ \ \ \ \ \ \ } \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x^{2} = 9 \\ y = \frac{2}{3}x \\ \end{matrix}\text{\ \ \ \ \ \ } \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = \pm 3 \\ y = \frac{2}{3}x \\ \end{matrix} \right.\ \]

\[Ответ:(3;2),\ ( - 3;\ - 2).\]