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МатематикаРешите систему уравнений x^2+2xy-3y^2=0; 2x^2+y^2=3.
Ответ:
\[\left\{ \begin{matrix}
x^{2} + 2xy - 3y^{2} = 0 \\
{2x}^{2} + y^{2} = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\end{matrix} \right.\ \]
\[\left( x^{2} + 2xy + y^{2} \right) - 4y^{2} = 0\]
\[(x + y)^{2} - (2y)^{2} = 0\]
\[(x + y - 2y)(x + y + 2y) = 0\]
\[(x - y)(x + 3y) = 0\]
\[x = y;\ \ x = - 3y.\]
\[1)\ x = y:\]
\[2y^{2} + y^{2} = 3\]
\[3y^{2} = 3\]
\[y^{2} = 1\]
\[y = \pm 1\]
\[\left\{ \begin{matrix}
y = 1 \\
x = 1 \\
\end{matrix}\ \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix}
y = - 1 \\
x = - 1 \\
\end{matrix} \right.\ \]
\[2)\ x = - 3y:\]
\[2 \cdot ( - 3y)^{2} + y^{2} = 3\]
\[18y^{2} + y^{2} = 3\]
\[19y^{2} = 3\]
\[y^{2} = \frac{3}{19}\]
\[y = \pm \frac{\sqrt{3}}{\sqrt{19}} = \frac{\sqrt{57}}{19}\]
\[x = - 3 \cdot \frac{\sqrt{57}}{19} = - \frac{3\sqrt{57}}{19}\]
\[x = - 3 \cdot \left( - \frac{\sqrt{57}}{19} \right) = \frac{3\sqrt{57}}{19}\]
\[\left\{ \begin{matrix}
y = - \frac{\sqrt{57}}{19}\text{\ \ } \\
x = - \frac{3\sqrt{57}}{19} \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix}
y = \frac{\sqrt{57}}{19}\text{\ \ } \\
x = \frac{3\sqrt{57}}{19} \\
\end{matrix} \right.\ \]
\[Ответ:(1;1);( - 1; - 1);\]
\[\left( - \frac{3\sqrt{57}}{19}; - \frac{\sqrt{57}}{19} \right);\left( \frac{3\sqrt{57}}{19};\frac{\sqrt{57}}{19} \right).\]
