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

Ответ:

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

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

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

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

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

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

\[1)\ x = y:\]

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

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

\[y^{2} = 1\]

\[y = \pm 1\]

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

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

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

\[18y^{2} + y^{2} = 3\]

\[19y^{2} = 3\]

\[y^{2} = \frac{3}{19}\]

\[y = \pm \frac{\sqrt{3}}{\sqrt{19}} = \frac{\sqrt{57}}{19}\]

\[x = - 3 \cdot \frac{\sqrt{57}}{19} = - \frac{3\sqrt{57}}{19}\]

\[x = - 3 \cdot \left( - \frac{\sqrt{57}}{19} \right) = \frac{3\sqrt{57}}{19}\]

\[\left\{ \begin{matrix} y = - \frac{\sqrt{57}}{19}\text{\ \ } \\ x = - \frac{3\sqrt{57}}{19} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = \frac{\sqrt{57}}{19}\text{\ \ } \\ x = \frac{3\sqrt{57}}{19} \\ \end{matrix} \right.\ \]

\[Ответ:(1;1);( - 1; - 1);\]

\[\left( - \frac{3\sqrt{57}}{19}; - \frac{\sqrt{57}}{19} \right);\left( \frac{3\sqrt{57}}{19};\frac{\sqrt{57}}{19} \right).\]