Решите систему уравнений x^2-2xy-3y^2=0; x^2+2y^2=3.

\[\left\{ \begin{matrix}
x^{2} - 2xy - 3y^{2} = 0 \\
x^{2} + 2y^{2} = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\end{matrix} \right.\ \ ( + )\]

\[\left( x^{2} + 2xy + y^{2} \right) - 4y^{2} = 0\]

\[(x + y)^{2} - (2y)^{2} = 0\]

\[(x + y - 2y)(x + y + 2y) = 0\]

\[(x - y)(x + 3y) = 0\]

\[x = y;\ \ \ x = - 3y:\]

\[1)\ x = y:\]

\[y^{2} + 2y^{2} = 3\]

\[3y^{2} = 3\]

\[y^{2} = 1\]

\[y = \pm 1\]

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

\[2)\ x = - 3y:\]

\[( - 3y)^{2} + 2y^{2} = 3\]

\[9y^{2} + 2y^{2} = 3\]

\[11y^{2} = 3\]

\[y^{2} = \frac{3}{11}\]

\[y = \pm \frac{\sqrt{3}}{\sqrt{11}} = \pm \frac{\sqrt{33}}{11}\]

\[\left\{ \begin{matrix}
y = \frac{\sqrt{33}}{11}\text{\ \ \ \ \ \ \ } \\
x = - \frac{3\sqrt{33}}{11} \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix}
y = - \frac{\sqrt{33}}{11} \\
x = \frac{3\sqrt{33}}{11}\text{\ \ } \\
\end{matrix} \right.\ \]

\(Ответ:( - 1;\ - 1);(1;1);\)

\[\left( - \frac{3\sqrt{33}}{11};\frac{\sqrt{33}}{11} \right);\left( \frac{3\sqrt{33}}{11};\ - \frac{\sqrt{33}}{11} \right).\]