ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 367

\[\boxed{\mathbf{367}\mathbf{.}}\]

\[4^{x}\left( \sqrt{16^{1 - x} - 1} + 2 \right) < 4\left| 4^{x} - 1 \right|\]

\[1)\ 4^{x} - 1 \geq 0\]

\[4^{x} \geq 1\]

\[4^{x} \geq 4^{0}\ \]

\[x \geq 0.\]

\[Если\ x \geq 0:\]

\[4^{x}\sqrt{\frac{16}{16^{x}} - 1} + 2 \bullet 4^{x} < 4 \bullet 4^{x} - 4\]

\[4^{x}\sqrt{\frac{16}{4^{2x}} - 1} < 2 \bullet 4^{x} - 4\]

\[16 - 4^{2x} < 4 \bullet 4^{2x} - 16 \bullet 4^{x} + 16\]

\[5 \bullet 4^{2x} - 16 \bullet 4^{x} > 0\]

\[4^{x} \bullet \left( 5 \bullet 4^{x} - 16 \right) > 0\]

\[5 \bullet 4^{x} - 16 > 0\]

\[5 \bullet 4^{x} > 16\]

\[4^{x} > \frac{16}{5}\ \ \ \ \ |\ :4^{2}\]

\[4^{x - 2} > \frac{1}{5}\]

\[\log_{4}4^{x - 2} > \log_{4}\left( \frac{1}{5} \right)\]

\[x - 2 > \log_{4}5^{- 1}\]

\[x > 2 - \log_{4}5.\]

\[Если\ x < 0:\]

\[4^{x}\sqrt{\frac{16}{16^{x}} - 1} + 2 \bullet 4^{x} < 4 - 4 \bullet 4^{x}\]

\[4^{x}\sqrt{\frac{16}{4^{2x}} - 1} < 4 - 6 \bullet 4^{x}\]

\[37 \bullet 4^{2x} - 48 \bullet 4^{x} > 0\]

\[4^{x} \bullet \left( 37 \bullet 4^{x} - 48 \right) > 0\]

\[37 \bullet 4^{x} - 48 > 0\]

\[37 \bullet 4^{x} > 48\]

\[4^{x} > \frac{48}{37}\]

\[\frac{48}{37} > 1 \Longrightarrow нет\ корней.\]

\[имеет\ смысл\ при:\]

\[16^{1 - x} - 1 \geq 0\]

\[16^{1 - x} \geq 1\]

\[16^{1 - x} \geq 16^{0}\]

\[1 - x \geq 0\ \]

\[x \leq 1.\]

\[Ответ:\ \ \ 2 - \log_{4}5 < x \leq 1.\]