\(\dfrac{{3{{\lg }^2}x-8}}{{{{\lg }^2}x-4}} \ge 2.\)
Пусть \(\lg x = t\). Тогда:
\(\dfrac{{3{t^2}-8}}{{{t^2}-4}} \ge 2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{t^2}}}{{{t^2}-4}} \ge 0.\)
Решим полученное неравенство методом интервалов. Нули числителя:
\({t^2} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,t = 0.\)
Нули знаменателя:
\({t^2}-4 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = -2,}\\{t = 2.\,\,}\end{array}} \right.\)
Следовательно: \(\left[ {\begin{array}{*{20}{c}}{t < -2,}\\{t = 0,\,\,}\\{t > 2.\,\,}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\,\begin{array}{*{20}{c}}{\lg x < -2,}\\{\lg x = 0,\,\,\,}\\{\lg x > 2\,\,\,\,}\end{array}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,} \right.\left[ {\begin{array}{*{20}{c}}{\lg x < \lg \frac{1}{{100}},}\\{\lg x = \lg 1,\,\,\,\,\,\,\,\,}\\{\lg x > \lg 100\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{0 < x < \dfrac{1}{{100}},}\\{x = 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x > 100\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,} \right.} \right.x\, \in \,\left( {0;\dfrac{1}{{100}}} \right) \cup \left\{ 1 \right\} \cup \left( {100;\infty } \right).\)
Ответ: \(\left( {0;\dfrac{1}{{100}}} \right) \cup \left\{ 1 \right\} \cup \left( {100;\infty } \right).\)