Запишем ОДЗ:
\(\left\{ \begin{array}{l}0,2 \cdot {5^x} > 0,\\{\log _3}\left( {0,2 \cdot {5^x}} \right) \ne 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ \begin{array}{l}{5^x} > 0,\\0,2 \cdot {5^x} \ne 1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ \begin{array}{l}x \in R,\\x \ne 1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x \ne 1.\)
\(\dfrac{{{x^2} + 4x-5}}{{{{\log }_3}\left( {0,2 \cdot {5^x}} \right)}} \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\dfrac{{{x^2} + 4x-5}}{{{{\log }_3}{5^{x-1}}}} \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\dfrac{{{x^2} + 4x-5}}{{\left( {x-1} \right){{\log }_3}5}} \ge 0.\)
Так как \({\log _3}5 > {\log _3}1 = 0,\) то последнее неравенство примет вид:
\(\dfrac{{{x^2} + 4x-5}}{{x-1}} \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\dfrac{{\left( {x + 5} \right)\left( {x-1} \right)}}{{x-1}} \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ \begin{array}{l}x + 5 \ge 0,\\x-1 \ne 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\left\{ \begin{array}{l}x \ge -5,\\x \ne 1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x \in \left[ {-5;1} \right) \cup \left( {1; + \infty } \right).\)
Так как ОДЗ \(x \ne 1,\) то решением исходного неравенства является: \(x \in \left[ {-5;1} \right) \cup \left( {1; + \infty } \right).\)
Ответ: \(\left[ {-5;1} \right) \cup \left( {1; + \infty } \right).\)