\(\dfrac{{5{{\lg }^2}x-1}}{{{{\lg }^2}x-1}} \ge 1.\)
Пусть \(\lg x = t\). Тогда:
\(\dfrac{{5{t^2}-1}}{{{t^2}-1}} \ge 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{4{t^2}}}{{{t^2}-1}} \ge 0.\)
Решим полученное неравенство методом интервалов. Нули числителя:
\(4{t^2} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,t = 0.\)
Нули знаменателя:
\({t^2}-1 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = -1,}\\{t = 1.\,\,}\end{array}} \right.\)
Следовательно: \(\left[ {\begin{array}{*{20}{c}}{t < -1,}\\{t = 0,\,\,}\\{t > 1.\,\,}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{\lg x < -1,}\\{\lg x = 0,\,\,}\\{\lg x > 1\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\lg x < \lg \frac{1}{{10}},}\\{\lg x = \lg 1,\,\,\,\,\,}\\{\lg x > \lg 10\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{0 < x < \dfrac{1}{{10}},}\\{x = 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x > 10\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\left( {0;\dfrac{1}{{10}}} \right) \cup \left\{ 1 \right\} \cup \left( {10;\infty } \right).\)
Ответ: \(\left( {0;\dfrac{1}{{10}}} \right) \cup \left\{ 1 \right\} \cup \left( {10;\infty } \right)\).