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Выражение имеет смысл при:

$x > 0$

$\lg^2x - 4 \neq 0 \implies \lg x \neq \pm 2 \implies x \neq 100, \ x \neq 0,01$

Сделаем замену $t = \lg x$ :

$\frac{4t^2 - 4}{t^2 - 4} - 1 \geq 0 \Rightarrow \frac{4t^2 - 4 - t^2 + 4}{t^2 - 4} \geq 0 \Rightarrow \frac{3t^2}{t^2 - 4} \geq 0 \Rightarrow \frac{t^2}{(t-2)(t+2)} \geq 0$

Воспользуемся методом интервалов

Тогда получаем

$t \in (-\infty; -2) \cup \{0\} \cup (2; +\infty)$

$\lg x < -2 \implies 0 < x < 0,01$
$\lg x = 0 \implies x = 1$
$\lg x > 2 \implies x > 100$

$\textbf{Ответ:} \ x \in (0; 0,01) \cup \{1\} \cup (100; +\infty)$

48:{"original_url":"https://storage.yandexcloud.net/egehub/public/776805/xROJAaiT4TVwNm1CfpuKJtAEq4ygpFHyqpoUsCEs.png","small_url

Задание №38965 ЕГЭ по Математике