\(\dfrac{4}{{{{\log }_2}x + 1}} \le 1.\)
Пусть \({\log _2}x = t\). Тогда:
\(\dfrac{4}{{t + 1}} \le 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{4}{{t + 1}}-1 \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{3-t}}{{t + 1}} \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t < -1,}\\{t \ge 3.\,\,}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{{{\log }_2}x < -1,}\\{{{\log }_2}x \ge 3\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\log }_2}x < {{\log }_2}\dfrac{1}{2},}\\{{{\log }_2}x \ge {{\log }_2}8\,\,}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{0 < x < \dfrac{1}{2},}\\{x \ge 8\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,} \right.} \right.\,x \in \,\left( {0;\dfrac{1}{2}} \right) \cup \left[ {8;\infty } \right).\)
Ответ: \(\left( {0;\dfrac{1}{2}} \right) \cup \left[ {8;\infty } \right).\)