Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 10

\[\boxed{\mathbf{10.}}\]

\[\textbf{а)}\ \sqrt{\log_{2}^{2}x + 3} = \log_{2}x - 1\]

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

\[\log_{2}^{2}x + 3 = \log_{2}^{2}x - 2\log_{2}x + 1\]

\[2\log_{2}x = - 2\]

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

\[x = 2^{- 1}\]

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

\[Проверка:\]

\[\sqrt{\log_{2}^{2}\frac{1}{2} + 3} = \log_{2}\frac{1}{2} - 1\]

\[\sqrt{\log_{2}^{2}\frac{1}{2} + 3} = - 1 - 1\]

\[\sqrt{\log_{2}^{2}\frac{1}{2} + 3} = - 2\]

\[Не\ может\ быть\ отрицательным;\]

\[нет\ корней.\]

\[Ответ:нет\ корней.\]

\[\textbf{б)}\ \sqrt{\log_{3}^{2}x + 5} = 1 - \log_{3}x\]

\[\log_{3}^{2}x + 5 = \left( 1 - \log_{3}x \right)^{2}\]

\[\log_{3}^{2}x + 5 = 1 - 2\log_{3}x + \log_{3}^{2}x\]

\[2\log_{3}x = - 4\]

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

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

\[x = \frac{1}{9}.\]

\[Проверка:\]

\[\sqrt{\log_{3}^{2}\frac{1}{9} + 5} = 1 - \log_{3}\frac{1}{9}\]

\[\sqrt{4\log_{3}^{2}3 + 5} = 1 + 2\]

\[\sqrt{4 + 5} = 3\]

\[\sqrt{9} = 3\]

\[3 = 3\]

\[x = \frac{1}{9} - корень.\]

\[Ответ:x = \frac{1}{9}.\]

\[\textbf{в)}\ \sqrt{4^{x + 1} - 2^{x + 1} - 3} = 2^{x} + 1\]

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

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

\[= 4^{x} + 2 \cdot 2^{x} + 1\]

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

\[t = 2^{x}:\]

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

\[D_{1} = 4 + 12 = 16\]

\[t_{1} = \frac{2 + 4}{3} = 2;\]

\[t_{2} = \frac{2 - 4}{3} = - \frac{2}{3}.\]

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

\[нет\ корней.\]

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

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

\[x = 1.\]

\[Проверка:\]

\[\sqrt{4^{2} - 2^{2} - 3} = 2 + 1\]

\[\sqrt{16 - 4 - 3} = 3\]

\[\sqrt{9} = 3\]

\[3 = 3\]

\[x = 1 - корень.\]

\[Ответ:x = 1.\]

\[\textbf{г)}\ \sqrt{3 \cdot 4^{x} - 2^{x} + 2} = 2^{x} + 1\]

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

\[3 \cdot 4^{x} - 2^{x} + 2 = 4^{x} + 2 \cdot 2^{x} + 1\]

\[2 \cdot 4^{x} - 3 \cdot 2^{x} + 1 = 0\]

\[t = 2^{x}:\]

\[2t^{2} - 3t + 1 = 0\]

\[D = 9 - 8 = 1\]

\[t_{1} = \frac{3 + 1}{4} = 1;\]

\[t_{2} = \frac{3 - 1}{4} = \frac{1}{2}.\]

\[1)\ 2^{x} = 1\]

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

\[x = 0.\]

\[2)\ 2^{x} = \frac{1}{2}\]

\[2^{x} = 2^{- 1}\]

\[x = - 1.\]

\[Проверка:\]

\[\sqrt{3 \cdot 4^{0} - 2^{0} + 2} = 2^{0} + 1\]

\[\sqrt{3 - 1 + 2} = 1 + 1\]

\[\sqrt{4} = 2\]

\[2 = 2\]

\[x = 0 - корень.\]

\[\sqrt{3 \cdot 4^{- 1} - 2^{- 1} + 2} = 2^{- 1} + 1\]

\[\sqrt{\frac{3}{4} - \frac{1}{2} + 2} = \frac{1}{2} + 1\ \]

\[\sqrt{2 + \frac{1}{4}} = 1\frac{1}{2}\]

\[\sqrt{\frac{9}{4}} = \frac{3}{2}\]

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

\[x = - 1 - корень.\]

\[Ответ:x = - 1;\ \ x = 0.\]

\[\textbf{д)}\ \sqrt{1 - \cos x} = \sin x\]

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

\[1 - \cos x = 1 - \cos^{2}x\]

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

\[\cos x\left( \cos x - 1 \right) = 0\]

\[1)\cos x = 0\]

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

\[2)\cos x - 1 = 0\]

\[\cos x = 1\]

\[x = 2\pi k.\]

\[Проверка:\]

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

\[\sqrt{1 - 0} = 1\]

\[1 = 1.\]

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

\[\sqrt{1 - 0} = - 1\]

\[\sqrt{1} = - 1\]

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

\[корнем.\]

\[\sqrt{1 - 1} = 0\]

\[0 = 0\]

\[x = 2\pi k - корень.\]

\[Ответ:x = \frac{\pi}{2} + 2\pi k;x = 2\pi k.\]

\[\textbf{е)}\ \sqrt{1 + \sin x} = \cos x\ \]

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

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

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

\[\sin x\left( \sin x + 1 \right) = 0\]

\[1)\sin x = 0\]

\[x = \pi k.\]

\[2)\sin x + 1 = 0\]

\[\sin x = - 1\]

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

\[Проверка:\]

\[\sqrt{1 + 0} = 1\]

\[1 = 1\]

\[x = 2\pi k - корень.\]

\[\sqrt{1 - 0} = - 1\]

\[\sqrt{1} \neq - 1\]

\[x = \pi + 2\pi k - не\ является\ \]

\[корнем.\]

\[\sqrt{1 - 1} = 0\]

\[0 = 0\]

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

\[Ответ:x = - \frac{\pi}{2} + 2\pi k;x = 2\pi k.\]