Решите систему уравнений: 2x^2+y^2=54; xy=-10.

Ответ:

\[\left\{ \begin{matrix} 2x^{2} + y^{2} = 54 \\ xy = - 10\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2x^{2} + y^{2} = 54 \\ y = - \frac{10}{x}\text{\ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2x² + \left( \frac{10}{x} \right)^{2} = 54\ \ | \cdot x^{2}² \\ y = - \frac{10}{x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[2x^{4} - 54x^{2} + 100 = 0\ \ |\ :2\]

\[x^{4} - 27x^{2} + 50 = 0\]

\[Пусть\ x^{2} = t:\]

\[t^{2} - 27t + 50 = 0\]

\[t_{1} + t_{2} = 27,\ \ \ t_{1} \cdot t_{2} = 50\]

\[t_{1} = 25,\ \ t_{2} = 2\]

\[\left\{ \begin{matrix} x^{2} = 25 \\ x^{2} = 2\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = \pm 5\ \ \\ x = \pm \sqrt{2} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 5\ \ \ \\ y = - 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = - 5 \\ y = 2\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = \sqrt{2\ }\text{\ \ \ \ \ } \\ y = - 5\sqrt{2} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = - \sqrt{2} \\ y = 5\sqrt{2} \\ \end{matrix} \right.\ \]

\[Ответ:(5;\ - 2);( - 5;2);\]

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