Решебник по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 9

\[\boxed{\mathbf{9.}}\]

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

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

\[(x + 1)^{2} = 0\]

\[x = - 1.\]

\[Проверим:\]

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

\[= \log_{3}(0 + 1) = 0.\]

\[Ответ:x = - 1.\]

\[\left\{ \begin{matrix} \log_{4}{((x + 2)^{2} + 1)} = 0\ \ \ \ \ \ \\ \log_{5}\left( \left( x^{3} + 2x^{2} \right)^{2} + 1 \right) = 0 \\ \end{matrix} \right.\ \]

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

\[(x + 2)^{2} = 0\]

\[x = - 2.\]

\[Проверим:\]

\[Ответ:x = - 2.\]

\[\left\{ \begin{matrix} \log_{6}{\left( (x + 3)^{2} + 1 \right)\ } = 0 \\ \log_{\frac{2}{7}}\left( \sqrt[3]{x + 2} + 2 \right) = 0\ \ \ \\ \end{matrix} \right.\ \]

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

\[(x + 3)^{2} = 0\]

\[x = - 3.\]

\[Проверим:\]

\[\log_{\frac{2}{7}}\left( \sqrt[3]{- 3 + 2} + 2 \right) = \log_{\frac{2}{7}}1 = 0.\]

\[Ответ:x = - 3.\]

\[\left\{ \begin{matrix} \log_{8}\left( (x + 4)^{2} + 1 \right) = 0 \\ \log_{\frac{2}{9}}\left( \sqrt[5]{x - 28} + 3 \right) = 0 \\ \end{matrix} \right.\ \]

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

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

\[x = - 4.\]

\[Проверим:\]

\[\log_{\frac{2}{9}}\left( \sqrt[5]{- 4 - 28} + 3 \right) =\]

\[= \log_{\frac{2}{9}}( - 2 + 3) = \log_{\frac{2}{9}}1 = 0.\]

\[Ответ:x = - 4.\]