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а) $\lg^2 x^3 - 20 \lg \sqrt{x} + 1 = 0.$

Запишем ОДЗ:
$\left\{ \begin{array}{l} x^3 > 0, \\ \sqrt{x} > 0 \end{array} \right. \quad \Leftrightarrow \quad x > 0.$

$\lg^2 x^3 - 20 \lg \sqrt{x} + 1 = 0 \quad \Leftrightarrow \quad (3 \lg x)^2 - 10 \lg x + 1 = 0 \quad \Leftrightarrow \quad 9 \lg^2 x - 10 \lg x + 1 = 0.$

Пусть $\lg x = t$ . Тогда уравнение примет вид:

$9t^2 - 10t + 1 = 0 \quad \Leftrightarrow \quad \left[ \begin{array}{l} t = \dfrac{1}{9}, \\ t = 1. \end{array} \right.$

Вернёмся к прежней переменной:

$\left[ \begin{array}{l} \lg x = \dfrac{1}{9}, \\ \lg x = 1 \end{array} \right. \quad \Leftrightarrow \quad \left[ \begin{array}{l} x = \sqrt[9]{10}, \\ x = 10. \end{array} \right.$

б) Отберём корни, принадлежащие отрезку $[2;\, 10].$

Так как $\sqrt[9]{10} < \sqrt[9]{512} = \sqrt[9]{2^9}$ = 2, то $x = \sqrt[9]{10} \notin [2;\, 10].$

$x = 10 \in [2;\, 10].</s</article></body></html>$

Задание №28247 ЕГЭ по Математике