Решите систему уравнений: x^2+3xy-10y^2=0; x^2+2xy-y^2=28.

Ответ:

\[\left\{ \begin{matrix} x² + 3xy - 10y^{2} = 0 \\ x² + 2xy - y^{2} = 28\ \ \\ \end{matrix} \right.\ \]

\[x^{2} + 3xy - 10y^{2} = 0\]

\[D = 9y^{2} + 40y^{2} = 49y^{2}\]

\[x_{1} = \frac{- 3y + 7y}{2} = 2y;\ \ \]

\[x_{2} = \frac{- 3y - 7y}{2} = - 5y\]

\[Ответ:(4;2);( - 4;\ - 2);\]

\[\left( - 5\sqrt{2};\sqrt{2} \right);\left( 5\sqrt{2};\ - \sqrt{2} \right).\]