Задача 3. Решите систему неравенств \(\left\{ {\begin{array}{*{20}{c}} {\dfrac{{{2^{4x + 2}}}}{{{4^{x + 1}}}} > 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {1 + {{\log }_3}\left( {x-4} \right) \leqslant {{\log }_3}\left( {x + 21} \right)} \end{array}} \right.\)
Ответ
ОТВЕТ: \(\left( {4;\,16,5\,} \right].\)
Решение
Решим первое неравенство:
\(\dfrac{{{2^{4x + 2}}}}{{{4^{x + 1}}}} > 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{{2^{4x + 2}}}}{{{2^{2x + 2}}}} > 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{2^{2x}} > {2^0}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,2x > 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x > 0.\)
Решим второе неравенство:
\({\log _3}3 + {\log _3}\left( {x-4} \right) \le {\log _3}\left( {x + 21} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\log _3}\left( {3x-12} \right) \le {\log _3}\left( {x + 21} \right)\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{3x-12 \le x + 21,}\\{3x-12 > 0\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2x \le 33,}\\{3x > 12\,}\end{array}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \le 16,5,}\\{x > 4\,\,\,\,\,\,\,}\end{array}} \right.} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \,\left( {4;\,16,5} \right].\)
Найдем общее решение:
Таким образом, решение исходной системы неравенств: \(x \in \left( {4;16,5} \right].\)
Ответ: \(\left( {4;16,5} \right].\)