-2\end{array}\right\; \; \; \; \Rightarrow \; \; \; x\in [\; 2\; ;\, +\infty )" alt="\left\{\begin{array}{l}x^2-4\geq 0\\(\frac{1}{5})^{x+1}<5\end{array}\right\; \; \left\{\begin{array}{l}(x-2)(x+2)\geq 0\\5^{-x-1}<5\end{array}\right\; \; \left\{\begin{array}{l}x\in (-\infty ,-2\, ]\cup [\; 2,+\infty )\\-x-1<1\end{array}\right\\\\\\\left\{\begin{array}{l}x\in (-\infty ,-2\, ]\cup [\; 2,+\infty )\\x>-2\end{array}\right\; \; \; \; \Rightarrow \; \; \; x\in [\; 2\; ;\, +\infty )" align="absmiddle" class="latex-formula">