Вопрос:

Решите систему уравнений: 2x^2+xy-3y^2=3; x^2-4xy-3y^2=9.

Ответ:

\[\left\{ \begin{matrix} 2x² + xy - 3y^{2} = 3 \\ x² - 4xy - 3y^{2} = 9 \\ \end{matrix} \right.\ \]

\[Пусть\ \ \ \ \ y = tx:\]

\[\left\{ \begin{matrix} 2x^{2} + tx^{2} - 3t^{2}x^{2} = 3 \\ x^{2} - 4tx^{2} - 3t^{2}x^{2} = 9 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x^{2}\left( 2 + t - 3t^{2} \right) = 3\ \ \ \\ x^{2}\left( 1 - 4t - 3t^{2} \right) = 9 \\ \end{matrix} \right.\ |\ :\]

\[\frac{2 + t - 3t^{2}}{1 - 4t - 3t^{2}} = \frac{1}{3}\]

\[6 + 3t - 9t^{2} - 1 + 4t + 3t^{2} = 0\]

\[- 6t^{2} + 7t + 5 = 0\]

\[D = 49 + 120 = 169\]

\[t_{1} = \frac{- 7 + 13}{- 12} = - \frac{1}{2}\]

\[t_{2} = \frac{- 7 - 13}{- 12} = \frac{5}{3}\]

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

\[\left\{ \begin{matrix} y = \frac{5}{3}\text{x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 2x² + \frac{5}{3}x² - \frac{25}{3}x^{2} = 3 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = - \frac{1}{2}\text{x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 8x^{2} - 2x^{2} - 3x^{2} = 12 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = \frac{5}{3}\text{x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 6x² + 5x² - 25x^{2} = 9 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = - \frac{1}{2}x \\ 3x^{2} = 12 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = \frac{5}{3}\text{x\ \ \ \ \ } \\ - 14x^{2} = 9 \\ \end{matrix} \right.\ \Longrightarrow нет\ корней.\]

\[\left\{ \begin{matrix} y = - \frac{1}{2}x \\ x = \pm 2\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} x = 2\ \ \ \\ y = - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]

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

\[Ответ:(2;\ - 1);( - 2;1)\text{.\ }\]

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