Решите систему уравнений: x^2+4xy+4y^2=1; 2x^2-3xy+y^2=6.

Ответ:

\[\left\{ \begin{matrix} x^{2} + 4xy + 4y^{2} = 1 \\ 2x^{2} - 3xy + y^{2} = 6 \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} (x + 2y)^{2} = 1\ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{2} - 3xy + y^{2} = 6 \\ \end{matrix} \right.\ \]

\[1.\ \left\{ \begin{matrix} x + 2y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{2} - 3xy + y^{2} = 6 \\ \end{matrix}\text{\ \ \ \ \ \ \ } \right.\ \]

\[15y^{2} - 11y - 4 = 0\]

\[D = 121 + 240 = 361\]

\[y = \frac{11 + 19}{30} = 1,\]

\[y = \frac{11 - 19}{30} = - \frac{4}{15}\]

\[\left\{ \begin{matrix} x = - 1 \\ y = 1\ \ \ \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ } \right.\ \text{\ \ \ }\left\{ \begin{matrix} x = \frac{23}{15}\text{\ \ \ \ } \\ y = - \frac{4}{15} \\ \end{matrix} \right.\ \]

\[2.\ \left\{ \begin{matrix} x + 2y = - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{2} - 3xy + y^{2} = 6 \\ \end{matrix}\text{\ \ \ \ \ \ \ } \right.\ \]

\[15y^{2} + 11y - 4 = 0\]

\[D = 121 + 240 = 361\]

\[y = \frac{- 11 + 19}{30} = \frac{4}{15}\]

\[y = \frac{- 11 - 19}{30} = - 1\]

\[\left\{ \begin{matrix} x = 1\ \ \ \\ y = - 1 \\ \end{matrix}\text{\ \ \ \ \ \ } \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} x = - \frac{23}{15} \\ y = \frac{4}{15}\text{\ \ \ \ } \\ \end{matrix} \right.\ \]

\[Ответ:( - 1;1),\ \left( 1\frac{8}{15};\ - \frac{4}{15} \right),\ \]

\[(1;\ - 1),\ \left( - 1\frac{8}{15};\frac{4}{15} \right)\text{.\ }\]