Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 375

\[\boxed{\mathbf{375}.}\]

\[\left\{ \begin{matrix} x^{2} - 6x - 3y - 1 = 0\ \ \ \\ y^{2} + 2x + 9y + 14 = 0 \\ \end{matrix} \right.\ \]

\[3y = x^{2} - 6x - 1\]

\[y = \frac{x^{2} - 6x - 1}{3}\]

\[\left( \frac{x^{2} - 6x - 1}{3} \right)^{2} + 2x +\]

\[+ 3 \cdot \left( x^{2} - 6x - 1 \right) +\]

\[+ 14 = 0\ \ \ | \cdot 9\]

\[\left( x^{2} - (6x + 1) \right)^{2} + 18x +\]

\[+ 27x^{2} - 162x - 27 + 126 = 0\]

\[x^{4} - 12x^{3} + 61x^{2} -\]

\[- 132x + 100 = 0\]

\[P(x) =\]

\[= (x - 2)^{2}\left( x^{2} - 8x + 25 \right) = 0\]

\[x^{2} - 8x + 25 = 0\]

\[D_{1} = 16 - 25 < 0\]

\[нет\ корней.\]

\[x = 2:\]

\[y = \frac{4 - 12 - 1}{3} = - \frac{9}{3} = - 3.\]

\[Ответ:(2;\ - 3).\]