\({\left( {\dfrac{6}{{11}}} \right)^{\dfrac{{5x + 1}}{x}-1}} \le \dfrac{{121}}{{36}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {\dfrac{6}{{11}}} \right)^{\dfrac{{5x + 1}}{x}-1}} \le {\left( {\dfrac{6}{{11}}} \right)^{-2}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{5x + 1}}{x}-1 \ge -2\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{5x + 1}}{x} + 1 \ge 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{6x + 1}}{x} \ge 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {-\infty ;-\dfrac{1}{6}} \right] \cup \left( {0;\infty } \right).\)
Ответ: \(\left( {-\infty ;-\dfrac{1}{6}} \right] \cup \left( {0;\infty } \right).\)