ГДЗ по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 12

\[\boxed{\mathbf{12.}}\]

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

\[\left\{ \begin{matrix} \log_{3}x + 1 = \log_{3}^{2}x - 5 \\ \log_{3}x + 1 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\log_{3}x \geq - 1.\]

\[\log_{3}^{2}x - \log_{3}x - 6 = 0\]

\[t = \log_{3}x:\]

\[t^{2} - t - 6 = 0\]

\[t_{1} + t_{2} = 1;\ \ t_{1} \cdot t_{2} = - 6\]

\[t_{1} = 3;\ \ \ \]

\[t_{2} = - 2 < - 1\ (не\ подходит).\]

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

\[x = 3^{3} = 27.\]

\[Ответ:x = 27.\]

\[\textbf{б)}\ \sqrt{2\log_{4}x} = \sqrt{\log_{4}^{2}x - 8}\]

\[\left\{ \begin{matrix} 2\log_{4}x = \log_{4}^{2}x - 8 \\ 2\log_{4}x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[2\log_{4}x \geq 0\]

\[\log_{4}x \geq 0.\]

\[\log_{4}^{2}x - 2\log_{4}x - 8 = 0\]

\[t = \log_{4}x:\]

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

\[D_{1} = 1 + 8 = 9\]

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

\[t_{2} = 1 - 3 =\]

\[= - 2 < 0\ (не\ подходит).\]

\[\log_{4}x = 4\]

\[x = 4^{4} = 256.\]

\[Ответ:x = 256.\]

\[\textbf{в)}\ \sqrt{1 - 4\log_{\frac{1}{2}}x} = \sqrt{6 - \log_{\frac{1}{2}}^{2}x}\]

\[\left\{ \begin{matrix} 1 - 4\log_{\frac{1}{2}}x = 6 - \log_{\frac{1}{2}}^{2}x \\ 1 - 4\log_{\frac{1}{2}}x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1 - 4\log_{\frac{1}{2}}x \geq 0\]

\[4\log_{\frac{1}{2}}x \leq 1\]

\[\log_{\frac{1}{2}}x \leq \frac{1}{4}.\]

\[\log_{\frac{1}{2}}^{2}x - 4\log_{\frac{1}{2}}x - 5 = 0\]

\[t = \log_{\frac{1}{2}}x:\]

\[t^{2} - 4t - 5 = 0\]

\[D_{1} = 4 + 5 = 9\]

\[t_{1} = 2 + 3 =\]

\[= 5 > \frac{1}{4}\ (не\ подходит);\]

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

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

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

\[x = 2.\]

\[Ответ:x = 2.\]

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

\[\left\{ \begin{matrix} 2 - \log_{\frac{1}{3}}x = \log_{\frac{1}{3}}^{2}x - 4 \\ 2 - \log_{\frac{1}{3}}x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[2 - \log_{\frac{1}{3}}x \geq 0\]

\[\log_{\frac{1}{3}}x \leq 2.\]

\[\log_{\frac{1}{3}}^{2}x + \log_{\frac{1}{3}}x - 6 = 0\]

\[\log_{\frac{1}{3}}x = t:\]

\[t^{2} + t - 6 = 0\]

\[t_{1} + t_{2} = - 1;\ \ t_{1} \cdot t_{2} = - 6\]

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

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

\[x = \left( \frac{1}{3} \right)^{- 3}\]

\[x = 27.\]

\[\log_{\frac{1}{3}}x = 2\]

\[x = \left( \frac{1}{3} \right)^{2}\]

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

\[Ответ:x = \frac{1}{9};\ \ x = 27.\]