Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 367

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\[1)\ \left\{ \begin{matrix} \frac{x^{3}}{y} + xy = 10 \\ \frac{y^{3}}{x} + xy = \frac{5}{2}\text{\ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{x^{3}}{y} = 10 - xy \\ \frac{y^{3}}{x} = \frac{5}{2} - xy\ \ \\ \end{matrix} \right.\ (*)\]

\[\frac{x^{3}}{y} \cdot \frac{y^{3}}{x} = (10 - xy)\left( \frac{5}{2} - xy \right)\]

\[x^{2} \cdot y^{2} = 25 - \frac{5}{2}xy -\]

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

\[\frac{25}{2}xy = 25\ \ \ \ \ \ | \cdot \frac{2}{25}\]

\[xy = 2\]

\[y = \frac{2}{x}.\]

\[1)\ xy = 2;\ \ y = \frac{2}{x}:\]

\[\frac{x^{3}}{y} + xy = 10\]

\[\frac{x^{4}}{2} + 2 = 10\]

\[\frac{x^{4}}{2} = 8\]

\[x^{4} = 16\]

\[x = \pm 2.\]

\[2)\ x = 2;\ \ y = \frac{2}{x} = 1;\]

\[x = - 2;\ \ \ \ y = \frac{2}{x} = - 1.\]

\[Ответ:( - 2;\ - 1);(2;1).\]

\[2)\ \left\{ \begin{matrix} \frac{x^{3}}{y} + xy = 5\ \ \\ \frac{y^{3}}{x} + xy = \frac{10}{3} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{x^{3}}{y} = 5 - xy\ \ \\ \frac{y^{3}}{x} = \frac{10}{3} - xy \\ \end{matrix} \right.\ (*)\]

\[\frac{x^{3}}{y} \cdot \frac{y^{3}}{x} = (5 - xy)\left( \frac{10}{3} - xy \right)\]

\[x^{2} \cdot y^{2} = \frac{50}{3} - \frac{10}{3}xy -\]

\[- 5xy + x^{2}y^{2}\]

\[\frac{25}{3}xy = \frac{50}{3}\ \ \ \ \ \ | \cdot \frac{3}{25}\]

\[xy = 2\]

\[y = \frac{2}{x}.\]

\[1)\ y = \frac{2}{x};\ \ \ xy = 2:\]

\[\ \frac{x^{3}}{y} + xy = 5\]

\[\frac{x^{4}}{2} + 2 = 5\]

\[\frac{x^{4}}{2} = 3\]

\[x^{4} = 6\]

\[x = \pm \sqrt[4]{6}.\]

\[2)\ x = \sqrt[4]{6};\ \ y = \frac{2}{\sqrt[4]{6}} = \frac{\sqrt[4]{6}}{3};\]

\[x = - \sqrt[4]{6};\ \ y = - \frac{2}{\sqrt[4]{6}} = - \frac{\sqrt[4]{6}}{3}.\]

\[Ответ:\left( \sqrt[4]{6};\frac{\sqrt[4]{6}}{3} \right);\ \ \]

\[\left( - \sqrt[4]{6}; - \frac{\sqrt[4]{6}}{3} \right)\text{.\ }\]