Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 940

\[1)\ \left\{ \begin{matrix} y + 5 = x^{2}\text{\ \ \ \ } \\ x^{2} + y^{2} = 25 \\ \end{matrix} \right.\ \]

\[y + 5 = x^{2}\]

\[y = x^{2} - 5.\]

\[x^{2} + y^{2} = 25\]

\[x^{2} + \left( x^{2} - 5 \right)^{2} - 25 = 0\]

\[x^{2} + x^{4} - 10x^{2} + 25 - 25 = 0\]

\[x^{4} - 9x^{2} = 0\]

\[x^{2} \bullet \left( x^{2} - 9 \right) = 0\]

\[(x + 3) \bullet x^{2} \bullet (x - 3) = 0\]

\[x_{1} = - 3;\ \ \ y_{1} = 9 - 5 = 4;\]

\[x_{2} = 0;\ \ \ \ \ \ y_{2} = - 5;\]

\[x_{3} = 3;\ \ \ \ \ \ y_{3} = 9 - 5 = 4.\]

\[Ответ:\ \ ( - 3;\ 4);\ (0;\ - 5);\ (3;\ 4).\]

\[2)\ \left\{ \begin{matrix} xy = 16 \\ \frac{x}{y} = 4\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[xy = 16\]

\[x = \frac{16}{y}.\]

\[\frac{x}{y} = 4\]

\[\frac{16}{y} = 4y\ \ \ \ \ | \bullet y\]

\[16 = 4y^{2}\]

\[y^{2} = 4\]

\[y = \pm 2;\]

\[x = \frac{16}{\pm 2} = \pm 8.\]

\[Ответ:\ \ ( - 8;\ - 2);\ (8;\ 2).\]

\[3)\ \left\{ \begin{matrix} x^{2} + 2y^{2} = 96 \\ x = 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[(2y)^{2} + 2y^{2} = 96\]

\[4y^{2} + 2y^{2} = 96\]

\[6y^{2} = 96\]

\[y^{2} = 16\]

\[y = \pm 4;\]

\[x = 2 \bullet ( \pm 4) = \pm 8.\]

\[Ответ:\ \ ( - 8;\ - 4);\ (8;\ 4).\]