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$\circledast5x-21\ne0\Leftrightarrow x\ne\frac{21}5$

Пусть $t=\frac{10}{5x-21}$

${(t+\frac1t)}^2-\frac{25}4\le0$

Применим $a^2-b^2=(a-b)(a+b)$

$(t+\frac{1}{t}-\frac{5}{2})(t+\frac{1}{t}+\frac{5}{2})\leq0\Leftrightarrow\left[\begin{matrix}t+\frac1t-\frac52=0\mid\cdot2t\\t+\frac1t+\frac52=0\mid\cdot2t\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}\begin{cases}2t^2+2-5t=0\\t\ne0\end{cases}\\\begin{cases}2t^2+2+5t=0\\t\ne0\end{cases}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}t_1=\frac12\\t_2=2\\t_3=-2\\t_4=-\frac12\end{matrix}\right.$

Решим по методу интервалов: $t\in[-2;-\frac12]\cup[\frac12;2]$

Обратная замена

$-2\le\frac{10}{5x-21}\le-\frac12\Leftrightarrow\begin{cases}\frac{10}{5x-21}\le-\frac12\\\frac{10}{5x-21}\ge-2\end{cases}\Leftrightarrow\begin{cases}\frac{20+5x-21}{2(5x-21)}\le0\\\frac{10+10x-42}{5x-21}\ge0\end{cases}\Leftrightarrow\begin{cases}\frac{5x-1}{5x-21}\le0\\\frac{10x-32}{5x-21}\ge0\end{cases}\Leftrightarrow x\in[\frac15;\frac{16}5]$ $\be</article></body></html>$

Задание №4190 ЕГЭ по Математике