Решите систему уравнений: 4x^2-4xy+y^2=9; 3x^2+2xy-y^2=36.

Ответ:

\[\left\{ \begin{matrix} 4x^{2} - 4xy + y^{2} = 9\ \ \\ 3x^{2} + 2xy - y^{2} = 36 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} (2x - y)^{2} = 9\ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x² + 2xy - y^{2} = 36 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2x - y = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x^{2} + 2xy - y^{2} = 36 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2x - y = - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x² + 2xy - y^{2} = 36 \\ \end{matrix} \right.\ \]

\[3x² + 6x - 45 = 0\ \ \ |\ :3\]

\[x^{2} + 2x - 15 = 0\]

\[x_{1} + x_{2} = - 2,\ \ \ x_{1} \cdot x_{2} = - 15\]

\[x_{1} = - 5,\ \ x_{2} = 3\]

\[\left\{ \begin{matrix} x = - 5\ \ \ \\ y = - 13 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \left\{ \begin{matrix} x = 3 \\ y = 3 \\ \end{matrix} \right.\ \]

\[3x² - 6x - 45 = 0\ \ \ |\ :3\]

\[x^{2} - 2x - 15 = 0\]

\[x_{1} + x_{2} = 2,\ \ \ x_{1} \cdot x_{2} = - 15\]

\[x_{1} = 5,\ \ x_{2} = - 3\]

\[\left\{ \begin{matrix} x = 5\ \ \\ y = 13 \\ \end{matrix} \right.\ \ \ \ \ \ \ \ \ или\ \ \ \ \left\{ \begin{matrix} x = - 3 \\ y = - 3 \\ \end{matrix} \right.\ \]

\[Ответ:( - 5;\ - 13);\ \ (3;3);\ \ \]

\[(5;13);\ \ ( - 3;\ - 3).\]