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Напишем ОДЗ:

$25x^2-10x-8\neq0\Leftrightarrow\begin{cases}x\neq-0,4\\x\neq0,8&\end{cases}$

Введем замену:

$t=\frac2{25x^2-10x-8}$ $\left(t+\frac1t\right)^2\geq4\Leftrightarrow\left(\frac{t^2+1}t\right)^2\geq4\Leftrightarrow\left(\frac{t^2+1}t\right)^2-4\geq0\Leftrightarrow\left(\frac{t^2+1}t-2\right)\left(\frac{t^2+1}t+2\right)\geq0$

Найдем нули данной функции:

$\left[\begin{matrix}\frac{t^2+1}t-2=0\\\frac{t^2+1}t+2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}\frac{t^2-2t+1}t=0\\\frac{t^2+2t+1}t=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}\frac{(t-1)^2}t=0\\\frac{(t+1)^2}t=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}t=1-\text{корень четной степени}\\t=0-\text{корень четной степени}\\t=-1-\text{корень четной степени}\end{matrix}\right.$

Нанесем корни на интервал и найдем ответ

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

Обратная замена:

$t\neq0\Leftrightarrow\frac2{25x^2-10x-8}\neq0$

Др

Задание №3788 ЕГЭ по Математике