\(\dfrac{1}{{6{x^2}-5x}} \ge \dfrac{1}{{\sqrt {6{x^2}-5x + 1} -1}}\)
Учтём, что: \(6{x^2}-5x + 1 \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\, \in \,\left( {-\infty ;\dfrac{1}{3}} \right] \cup \left[ {\dfrac{1}{2};\infty } \right).\)
\(\dfrac{1}{{6{x^2}-5x}} \ge \dfrac{1}{{\sqrt {6{x^2}-5x + 1} -1}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{1}{{6{x^2}-5x + 1-1}} \ge \dfrac{1}{{\sqrt {6{x^2}-5x + 1} -1}}.\)
Пусть \(\sqrt {6{x^2}-5x + 1} = t,\) где \(t \ge 0.\) Тогда: \(6{x^2}-5x + 1 = {t^2}.\)
\(\left\{ \begin{array}{l}t \ge 0,\\\dfrac{1}{{{t^2}-1}} \ge \dfrac{1}{{t-1}}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}t \ge 0,\\\dfrac{1}{{\left( {t-1} \right)\left( {t + 1} \right)}}-\dfrac{1}{{t-1}} \ge 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}t \ge 0,\\\dfrac{t}{{\left( {t-1} \right)\left( {t + 1} \right)}} \le 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}t \in \left[ {0;\infty } \right),\\t \in \left( {-\infty ;-1} \right) \cup \left[ {0;1} \right)\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,t \in \left[ {0;1} \right).\)
Возвращаясь к прежней переменной, получим:
\(0 \le \sqrt {6{x^2}-5x + 1} < 1\,\,\,\,\, \Leftrightarrow \,\,\,\,\left\{ \begin{array}{l}6{x^2}-5x + 1 < 1,\\6{x^2}-5x + 1 \ge 0\end{array} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\left\{ \begin{array}{l}x \in \left( {0;\dfrac{5}{6}} \right),\\x\, \in \,\left( {-\infty ;\dfrac{1}{3}} \right] \cup \left[ {\dfrac{1}{2};\infty } \right)\end{array} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,x \in \left( {0;\dfrac{1}{3}} \right] \cup \left[ {\dfrac{1}{2};\dfrac{5}{6}} \right).\)
Ответ: \(\left( {0;\dfrac{1}{3}} \right] \cup \left[ {\dfrac{1}{2};\dfrac{5}{6}} \right).\)