Решите систему уравнений: x^2-2xy-y^2=7; x^2+xy+8y^2=14.

Ответ:

\[\left\{ \begin{matrix} x^{2} - 2xy - y^{2} = 7\ \ \\ x^{2} + xy + 8y^{2} = 14 \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[Пусть\ y = kx:\]

\[\left\{ \begin{matrix} 3x^{2} - 2kx^{2} - k^{2}x^{2} = 7 \\ x^{2} + kx^{2} + 8k^{2}x^{2} = 14 \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x^{2}\left( 3 - 2k - k^{2} \right) = 7\ \ \ \\ x^{2}\left( 1 + k + 8k^{2} \right) = 14 \\ \end{matrix}\text{\ \ \ \ } \right.\ \]

\[\frac{3 - 2k - k^{2}}{1 + k + 8k^{2}} = \frac{1}{2}\]

\[6 - 4k - 2k^{2} = 1 + k + 8k^{2}\text{\ \ \ \ \ \ }\]

\[- 10k^{2} - 5k + 5 = 0\ \ \ |\ :5\]

\[- 2k^{2} - k + 1 = 0\ \ \ \ \]

\[k_{1} + k_{2} = - \frac{1}{2};\ \ \ \ \ \ \ \ k_{1} = \frac{1}{2}\]

\[k_{1} \cdot k_{2} = - \frac{1}{2};\ \ \ \ \ \ \ \ \ k_{2} = - 1\]