ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 961

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

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

\[\ln x^{3} = \ln y^{2}\]

\[3\ln x = 2\ln y\]

\[\frac{\ln x}{\ln y} = \frac{2}{3}.\]

\[x^{y} = y^{x}\]

\[\ln x^{y} = \ln y^{x}\]

\[y\ln x = x\ln y\]

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

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

\[2y = 3x\]

\[y = \frac{3x}{2}.\]

\[Подставим:\]

\[x^{3} = \frac{9x^{2}}{4}\]

\[4x^{3} = 9x^{2}\]

\[4x^{3} - 9x^{2} = 0\]

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

\[x_{1} = 0;\ \ \ x_{2} = \frac{9}{4};\]

\[y_{1} = \frac{3 \bullet 0}{2} = 0;\]

\[y_{2} = \frac{3}{2} \bullet \frac{9}{4} = \frac{27}{8}.\]

\[Одно\ из\ решений:\]

\[x = 1;\ \ \ y = 1.\]

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

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

\[x^{\sqrt{y}} = y\]

\[\ln x^{\sqrt{y}} = \ln y\]

\[\sqrt{y}\ln x = \ln y\]

\[\frac{\ln y}{\ln x} = \sqrt{y}.\]

\[y^{\sqrt{y}} = x^{4}\]

\[\ln y^{\sqrt{y}} = \ln x^{4}\]

\[\sqrt{y}\ln y = 4\ln x\]

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

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

\[y = 4.\]

\[Подставим:\]

\[x^{\sqrt{4}} = 4\]

\[x^{2} = 4\]

\[x = \pm 2.\]

\[Одно\ из\ решений:\]

\[x = 1;\ \ \ y = 1.\]

\[Ответ:\ \ (1;\ 1);\ (2;\ 4).\]

\[3)\ \left\{ \begin{matrix} \sqrt{2}\sin x = \sin y\text{\ \ \ \ \ \ \ } \\ \sqrt{2}\cos x = \sqrt{3}\cos y \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2\sin^{2}x = \sin^{2}y\text{\ \ \ \ } \\ 2\cos^{2}x = 3\cos^{2}y \\ \end{matrix}\ \ ( + ) \right.\ \]

\[2\sin^{2}x + 2\cos^{2}x = \sin^{2}y + 3\cos^{2}y\]

\[2 = 1 + 2\cos^{2}y\]

\[2\cos^{2}y = 1\]

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

\[1 + \cos{2y} = 1\]

\[\cos{2y} = 0\]

\[2y = \frac{\pi}{2} + \pi k\]

\[y = \frac{\pi}{4} + \frac{\text{πk}}{2}.\]

\[1)\ \sqrt{2}\sin x = \frac{\sqrt{2}}{2}\]

\[\sin x = \frac{1}{2}\]

\[x = \frac{\pi}{6} + 2\pi n.\]

\[\sqrt{2}\cos x = \sqrt{3} \bullet \frac{\sqrt{2}}{2}\]

\[\cos x = \frac{\sqrt{3}}{2}\text{\ \ \ }\]

\[y = \frac{\pi}{4} + 2\pi k.\]

\[2)\ \sqrt{2}\sin x = \frac{\sqrt{2}}{2}\]

\[\sin x = \frac{1}{2}\]

\[x = \frac{5\pi}{6} + 2\pi n.\]

\[\sqrt{2}\cos x = \sqrt{3}\left( - \frac{\sqrt{2}}{2} \right)\]

\[\cos x = - \frac{\sqrt{3}}{2}\text{\ \ }\]

\[y = \frac{3\pi}{4} + 2\pi k.\]

\[3)\ \sqrt{2}\sin x = - \frac{\sqrt{2}}{2}\]

\[\sin x = - \frac{1}{2}\]

\[x = \frac{7\pi}{6} + 2\pi n.\]

\[\sqrt{2}\cos x = \sqrt{3}\left( - \frac{\sqrt{2}}{2} \right)\]

\[\cos x = - \frac{\sqrt{3}}{2}\]

\[y = \frac{5\pi}{4} + 2\pi k.\]

\[4)\ \sqrt{2}\sin x = - \frac{\sqrt{2}}{2}\]

\[\sin x = - \frac{1}{2}\]

\[x = \frac{11\pi}{6} + 2\pi n.\]

\[\sqrt{2}\cos x = \sqrt{3} \bullet \frac{\sqrt{2}}{2}\]

\[\cos x = \frac{\sqrt{3}}{2}\]

\[y = \frac{7\pi}{4} + 2\pi k.\]

\[Ответ:\ \ \]

\[\left( \frac{\pi}{6} + 2\pi n;\ \frac{\pi}{4} + 2\pi k \right);\ \]

\[\left( \frac{5\pi}{6} + 2\pi n;\ \frac{3\pi}{4} + 2\pi k \right);\]

\[\left( \frac{7\pi}{6} + 2\pi n;\ \frac{5\pi}{4} + 2\pi k \right);\ \]

\[\left( \frac{11\pi}{6} + 2\pi n;\ \frac{7\pi}{4} + 2\pi k \right).\]

\[4)\ \left\{ \begin{matrix} x - y = - \frac{1}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \cos^{2}\text{πx} - \sin^{2}\text{πy} = \frac{1}{2} \\ \end{matrix} \right.\ \]

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

\[y = x + \frac{1}{3}.\]

\[\cos^{2}\text{πx} - \sin^{2}\text{πy} = \frac{1}{2}\]

\[\frac{1 + \cos{2\pi x}}{2} - \frac{1 - \cos{2\pi y}}{2} = \frac{1}{2}\]

\[\cos{2\pi x} + \cos{2\pi y} = 1\]

\[2\cos(\pi x + \pi y) \bullet \cos(\pi x - \pi y) = 1\]

\[\cos\left( \pi x + \pi x + \frac{\pi}{3} \right) \bullet \cos\left( \pi x - \pi x - \frac{\pi}{3} \right) = \frac{1}{2}\]

\[\cos\left( 2\pi x + \frac{\pi}{3} \right) \bullet \cos\frac{\pi}{3} = \frac{1}{2}\]

\[\cos\left( 2\pi x + \frac{\pi}{3} \right) \bullet \frac{1}{2} = \frac{1}{2}\]

\[\cos\left( 2\pi x + \frac{\pi}{3} \right) = 1\]

\[2\pi x + \frac{\pi}{3} = 2\pi n\]

\[2x + \frac{1}{3} = 2n\]

\[2x = 2n - \frac{1}{3}\]

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

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

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

\[5)\ \left\{ \begin{matrix} \cos x\sin y = \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ } \\ \sin{2x} + \sin{2y} = 0 \\ \end{matrix} \right.\ \]

\[\sin{2x} + \sin{2y} = 0\]

\[2\sin(x + y) \bullet \cos(x - y) = 0\]

\[\sin(x + y) = 0\]

\[x + y = \pi k\]

\[y = \pi k - x.\]

\[\cos(x - y) = 0\]

\[x - y = \frac{\pi}{2} + \pi k\]

\[y = x - \frac{\pi}{2} - \pi k;\]

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

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

\[0 - \sin(x - y) = 1\]

\[\sin(x - y) = - 1\]

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

\[x - \pi k + x = - \frac{\pi}{2} + 2\pi n\]

\[2x = - \frac{\pi}{2} + 2\pi n + \pi k\]

\[x = - \frac{\pi}{4} + \pi n + \frac{\text{πk}}{2};\]

\[y = \frac{\pi}{4} - \pi n + \frac{\text{πk}}{2}.\]

\[Корни\ совпадают:\]

\[\cos(x - y) = 0\]

\[\cos x\sin y = \frac{1}{2}.\]

\[\sin(x - y) = \pm 1\]

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

\[\frac{1}{2}\left( \sin(x + y) - ( \pm 1) \right) = \frac{1}{2}\]

\[\sin(x + y) \mp 1 = 1\]

\[\sin(x + y) = 0;\]

\[\sin(x + y) = 2.\]

\[Ответ:\ \ \]

\[\left( - \frac{\pi}{4} + \pi n + \frac{\text{πk}}{2};\ \frac{\pi}{4} - \pi n + \frac{\text{πk}}{2} \right).\]