Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1423

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

\[\left\{ \begin{matrix} y = x^{2} - 5\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + y^{2} - 25 = 0 \\ \end{matrix} \right.\ \]

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

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

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

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

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

\[x_{1} = - 3;\ \ x_{2} = 0;\ \ x_{3} = 3;\]

\[y_{1} = ( - 3)^{2} - 5 = 9 - 5 = 4;\]

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

\[y_{3} = (3)^{2} - 5 = 9 - 5 = 4.\]

\[Ответ:\ \ \]

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

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

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

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

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

\[y^{2} = 4\]

\[y = \pm 2;\]

\[x_{1} = 4 \bullet ( - 2) = - 8;\]

\[x_{2} = 4 \bullet 2 = 8.\]

\[Ответ:\ \ ( - 8;\ - 2);\ \ (8;\ 2).\]

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

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

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

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

\[y^{2} = 16\]

\[y = \pm 4;\]

\[x_{1} = 2 \bullet ( - 4) = - 8;\]

\[x_{2} = 2 \bullet 4 = 8.\]

\[Ответ:\ \ ( - 8;\ - 4);\ \ (8;\ 4).\]