\({\log _{\dfrac{1}{3}}}x + {\log _{\dfrac{1}{3}}}\left( {4-x} \right) > -1.\)
Найдём ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x > 0,\,\,\,\,\,}\\{4-x > 0}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x > 0,}\\{x < 4}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\left( {0;4} \right).} \right.} \right.\)
\({\log _{\dfrac{1}{3}}}x + {\log _{\dfrac{1}{3}}}\left( {4-x} \right) > {\log _{\dfrac{1}{3}}}3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _{\dfrac{1}{3}}}\left( {x \cdot \left( {4-x} \right)} \right) > {\log _{\dfrac{1}{3}}}3\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,4x-{x^2} < 3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{x^2}-4x + 3 > 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {-\infty ;1} \right) \cup \left( {3;\infty } \right).\)
Тогда общее решение с ОДЗ будет иметь вид:
\(\left\{ {\begin{array}{*{20}{c}}{x\, \in \,\left( {0;4} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x\, \in \,\left( {-\infty ;1} \right) \cup \left( {3;\infty } \right)}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x \in \,\left( {0;1} \right) \cup \left( {3;4} \right).} \right.\)
Ответ: \(\left( {0;1} \right) \cup \left( {3;4} \right).\)