Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1357

\[\boxed{\mathbf{1357}\mathbf{.}}\]

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

\[\log_{4}\left( 2 + \sqrt{x + 3} \right) = \log_{4}4\]

\[2 + \sqrt{x + 3} = 4\]

\[\sqrt{x + 3} = 2\]

\[x + 3 = 4\]

\[x = 1.\]

\[Ответ:\ \ x = 1.\]

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

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

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

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

\[D = 4 + 12 = 16\]

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

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

\[Ответ:\ \ x_{1} = - 1;\ \ x_{2} = 3.\]

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

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

\[\log_{3}(x + 1)^{\frac{1}{2}} =\]

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

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

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

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

\[4(x + 1) = x + 4\]

\[4x + 4 = x + 4\]

\[3x = 0\]

\[x = 0.\]

\[Ответ:\ \ x = 0.\]