Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 41

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\[\textbf{а)}\ \sqrt{\log_{2}(x + 2) + \log_{2}(x + 1)} =\]

\[= \sqrt{\log_{2}{(x - 2)} + \log_{2}{(2x - 1)}}\]

\[\log_{2}(x + 2) + \log_{2}(x + 1) =\]

\[= \log_{2}{(x - 2)} + \log_{2}{(2x - 1)}\]

\[\log_{2}\left( (x + 2)(x + 1) \right) =\]

\[= \log_{2}\left( (x - 2)(2x - 1) \right)\]

\[x > 2.\]

\[(x + 2)(x + 1) =\]

\[= (x - 2)(2x - 1)\]

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

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

\[x^{2} - 8x = 0\]

\[x(x - 8) = 0\]

\[x = 0 < 2 - не\ корень.\]

\[x = 8.\]

\[Ответ:x = 8.\]

\[\textbf{б)}\ \sqrt{\log_{3}(2x - 1) + \log_{3}(x - 4)} =\]

\[= \sqrt{2\log_{3}{(x - 2)}}\]

\[\log_{3}(2x - 1) + \log_{3}(x - 4) =\]

\[= 2\log_{3}{(x - 2)}\]

\[\log_{3}\left( (2x - 1)(x - 4) \right) =\]

\[= \log_{3}(x - 2)^{2}\]

\[x > 4.\]

\[(2x - 1)(x - 4) = (x - 2)^{2}\]

\[2x^{2} - x - 8x + 4 = x^{2} - 4x + 4\]

\[x^{2} - 5x = 0\]

\[x(x - 5) = 0\]

\[x = 0 < 4 - не\ корень;\]

\[x = 5.\]

\[Ответ:x = 5.\]

\[\textbf{в)}\ \sqrt{\log_{2}{\sin x} + 2} =\]

\[= \sqrt{1 - \log_{2}{\cos x}}\]

\[\log_{2}{\sin x} + 2 = 1 - \log_{2}{\cos x}\]

\[\log_{2}{\sin x - \log_{2}{\cos x}} = - 1\]

\[\log_{2}\left( \sin x\cos x \right) = \log_{2}\frac{1}{2}\]

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

\[2\sin x\cos x = 1\]

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

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

\[x = \frac{\pi}{4} + \pi n.\]

\[Проверка\ показала,\ что\ x =\]

\[= \frac{\pi}{4} + \pi n - корень\ уравнения.\]

\[Ответ:x = \frac{\pi}{4} + \pi n.\]

\[\textbf{г)}\ \sqrt{\log_{2}\left( - \cos x \right) + 3} =\]

\[= \sqrt{2 - \log_{2}\left( - \sin x \right)}\]

\[\log_{2}\left( - \cos x \right) + 3 =\]

\[= 2 - \log_{2}\left( - \sin x \right)\]

\[\log_{2}\left( - \cos x \right) + \log_{2}\left( - \sin x \right) =\]

\[= - 1\]

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

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

\[2\sin x\cos x = 1\]

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

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

\[x_{1} = \frac{\pi}{4} + \pi n;\]

\[x_{2} = - \frac{3\pi}{4} + 2\pi n.\]

\[Проверка\ показала,\ что\ корень\ \]

\[исходного\ уравнения\ x_{2}.\]

\[Ответ:x = - \frac{3\pi}{4} + 2\pi n.\]