Решите систему уравнений: x^2+6xy+9y^2=4; x^2-xy-4y^2=-2.

Ответ:

\[\left\{ \begin{matrix} x^{2} + 6xy + 9y^{2} = 4 \\ x^{2} - xy - 4y^{2} = - 2 \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ \ } \right.\ \]

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

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

\[8y^{2} - 14y + 6 = 0\ \ \ \ \ \ |\ :2\]

\[4y^{2} - 7y + 3 = 0\]

\[D = 49 - 48 = 1\]

\[y = \frac{7 + 1}{8} = 1\]

\[y = \frac{7 - 1}{8} = \frac{3}{4}\]

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

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

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

\[8y^{2} + 14y + 6 = 0\ \ \ \ \ \ \ |\ :2\]

\[4y^{2} + 7y + 3 = 0\]

\[D = 49 - 48 = 1\ \ \ \ \]

\[y_{1} = \frac{- 7 + 1}{8} = - \frac{3}{4},\ \ \ \ \ \ \]

\[y_{2} = \frac{- 7 - 1}{8} = - 1\]

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

\[Ответ:\ \ ( - 1;1),\ \left( - \frac{1}{4};\frac{3}{4} \right),\ \]

\[\left( \frac{1}{4};\ - \frac{3}{4} \right),\ (1;\ - 1).\]