(log по 5 (25х^2) + 48) / (log по 5 от х^2 - 49) >= -1

ОДЗ: x ≠ 0 (ОДЗ логарифма)
$\log_{5}{x^2}-49 \neq 0 \\ \log_{5}{x^2} \neq 49 \\ x^2 \neq 5^{49} \\ x \neq \pm5^{24}\sqrt{5}$

$\frac{\log_{5}{(25x^2)}+48}{\log_{5}{x^2}-49} \geq -1 \\ \frac{2\log_{5}{(|5x|)}+48}{2\log_{5}{|x|}-49} \geq -1 \\ \frac{2(\log_{5}{5}+\log_{5}{|x|})+48}{2\log_{5}{|x|}-49} \geq -1 \\ \frac{2(1+\log_{5}{|x|})+48}{2\log_{5}{|x|}-49} \geq -1 \\ \frac{2+2\log_{5}{|x|}+48}{2\log_{5}{|x|}-49} \geq -1 \\ \frac{2\log_{5}{|x|}+50}{2\log_{5}{|x|}-49} \geq -1$
Пусть $2\log_{5}{|x|}=t$
$\frac{t+50}{t-49} \geq -1 \\ \frac{t+50}{t-49} + 1 \geq 0 \\ \frac{2t+1}{t-49} \geq 0 \\ \left [ {{t \leq -0.5} \atop {t\ \textgreater \ 49}} \right. \\ \left [ {{2\log_{5}{|x| \leq -0.5}} \atop {2\log_{5}{|x|}\ \textgreater \ 49}} \right. \\ \left [ {{\log_{5}{|x|} \leq -0.25} \atop {\log_{5}{|x|}\ \textgreater \ \frac{49}{2} }} \right. \\ \left [ {{\log_{5}{|x|} \leq \log_{5}{5^{- \frac{1}{4} }}} \atop {\log_{5}{|x|}\ \textgreater \ \log_{5}{5^{\frac{49}{2}}} }} \right.$
$\left [ {{|x| \leq \frac{1}{ \sqrt[4]{5} } } \atop {|x|\ \textgreater \ \sqrt{5^{49}} }} \right. \\ \left [ { -\frac{ \sqrt[4]{125} }{5}\leq {x \leq \frac{ \sqrt[4]{125} }{5} } \atop { \left [ {{x\ \textless \ -5^{24} \sqrt{5} } \atop {x\ \textgreater \ 5^{24} \sqrt{5} }} \right. }} \right.$

Учитывая ОДЗ, получаем ответ:
$x\in(-\infty; -5^{24} \sqrt{5})\cup[-\frac{ \sqrt[4]{125} }{5}; 0)\cup(0; \frac{ \sqrt[4]{125} }{5}]\cup(5^{24}\sqrt{5}; +\infty)$