\(\left( {3-{{\log }_3}x} \right) \cdot {\log _5}\left( {x + 1} \right) \ge 0.\)
ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x + 1 > 0\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x > 0,\,}\\{x > -1}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\left( {0;\infty } \right).} \right.} \right.\)
Решим данное неравенство методом интервалов:
\(\left( {3-{{\log }_3}x} \right) \cdot {\log _5}\left( {x + 1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{3-{{\log }_3}x = 0,\,\,}\\{{{\log }_5}\left( {x + 1} \right) = 0}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 27,}\\{x = 0.\,\,\,}\end{array}} \right.} \right.\)
Таким образом, решением исходного неравенства является: \(x \in \,\left( {0;27} \right].\)
Ответ: \(\left( {0;27} \right].\)