ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1571

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\[\mathbf{Для\ первых\ двух\ систем}:\ \ \]

\[(1;\ 1).\]

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

\[\left\{ \begin{matrix} x^{y} = y^{x} \\ x = y^{\frac{2}{3}}\text{\ \ \ } \\ \end{matrix} \right.\ \]

\[\left( y^{\frac{2}{3}} \right)^{y} = y^{y^{\frac{2}{3}}}\]

\[y^{\frac{2y}{3}} = y^{y^{\frac{2}{3}}}\]

\[\frac{2y}{3} = y^{\frac{2}{3}}\]

\[y^{\frac{2}{3}} - \frac{2y}{3} = 0\]

\[y^{\frac{2}{3}} \bullet \left( 1 - \frac{2}{3}y^{\frac{1}{3}} \right) = 0\]

\[1)\ y^{\frac{2}{3}} = 0\]

\[y = 0;\]

\[x = 0^{\frac{2}{3}} = 0.\]

\[2)\ 1 - \frac{2}{3}y^{\frac{1}{3}} = 0\]

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

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

\[y = \left( \frac{3}{2} \right)^{3} = \frac{27}{8};\]

\[x = \left( \frac{27}{8} \right)^{\frac{2}{3}} = \left( \frac{3}{2} \right)^{2} = \frac{9}{4}.\]

\[Имеет\ смысл\ при:\]

\[\left\{ \begin{matrix} x \neq 0 \\ y \neq 0 \\ \end{matrix} \right.\ .\]

\[Ответ:\ \ \left( \frac{9}{4};\ \frac{27}{8} \right);\ \ (1;\ 1).\]

\[2)\ \left\{ \begin{matrix} x^{\sqrt{y}} = y\ \ \\ y^{\sqrt{y}} = x^{4} \\ \end{matrix} \right.\ \text{\ \ \ }\]

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

\[y = \left( y^{\frac{\sqrt{y}}{4}} \right)^{\sqrt{y}}\]

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

\[1 = \frac{y}{4}\]

\[y = 4;\]

\[x = 4^{\frac{\sqrt{4}}{4}} = 4^{\frac{2}{4}} = 4^{\frac{1}{2}} = 2.\]

\[Ответ:\ \ (2;\ 4);\ \ (1;\ 1).\]

\[3)\ \left\{ \begin{matrix} x - y = - \frac{1}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \cos^{2}\text{πx} - \sin^{2}\text{πy} = \frac{1}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x = y - \frac{1}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \cos^{2}\text{πx} - \sin^{2}\text{πy} = \frac{1}{2} \\ \end{matrix} \right.\ \]

\[\cos^{2}\left( \pi\left( y - \frac{1}{3} \right) \right) - \sin^{2}\text{πy} = \frac{1}{2}\]

\[\cos^{2}\left( \pi y - \frac{\pi}{3} \right) - \sin^{2}\text{πy} = \frac{1}{2}\]

\[\cos{2\pi y} + \sqrt{3}\sin{2\pi y} = 2\]

\[\frac{1}{2}\cos{2\pi y} + \frac{\sqrt{3}}{2}\sin{2\pi y} = 1\]

\[\sin\frac{\pi}{6} \bullet \cos{2\pi y} + \cos\frac{\pi}{6} \bullet \sin{2\pi y} = 1\]

\[\sin\left( \frac{\pi}{6} + 2\pi y \right) = 1\]

\[\frac{\pi}{6} + 2\pi y = \arcsin 1 + 2\pi n = \frac{\pi}{2} + 2\pi n\]

\[2\pi y = \frac{\pi}{2} - \frac{\pi}{6} + 2\pi n = \frac{\pi}{3} + 2\pi n\]

\[y = \frac{1}{2\pi} \bullet \left( \frac{\pi}{3} + 2\pi n \right) = \frac{1}{6} + n;\]

\[x = \frac{1}{6} + n - \frac{1}{3} = - \frac{1}{6} + n.\]

\[Ответ:\ \ \left( - \frac{1}{6} + n;\ \frac{1}{6} + n \right).\]