Задание №27649 ЕГЭ Профильной математике

Найдем ограничение на x:

$\left\{\begin{array}{l}x-5>0\\ 5^{2\left|x\right|+1}>0\\ 5^{\left|x\right|}>\frac{2}{11}\end{array}\right.\iff \left\{\begin{array}{l}x>5\\ x\in \mathrm{ℝ}\\ 5^{\left|x\right|}>5^{{\log }_5\frac{2}{11}}\end{array}\iff \right.\left\{\begin{array}{l}x>5\\ x\in \mathrm{ℝ}\end{array}\iff \right.x>5$

Разделим неравенство на $\sqrt{x-5}>0$:

${\log }_2\left(5^{2\left|x\right|+1}\right)-{\log }_2(11·5^{\left|x\right|}-2)>0$

Решим неравенство методом интервалов:

$$\begin{gathered} {\log }_2\left(5^{2\left|\mathrm{x}\right|+1}\right)-{\log }_2(11·5^{\left|\mathrm{x}\right|}-2) =0 \\ {\log }_2\left(5^{2\left|\mathrm{x}\right|+1}\right)={\log }_2(11·5^{\left|\mathrm{x}\right|}-2) \\ {5}^{2\left|\mathrm{x}\right|+1}=11·5^{\left|\mathrm{x}\right|}-2 \\ {5}^{2\left|\mathrm{x}\right|+1}-11·5^{\left|\mathrm{x}\right|}+2=0 \\ 5·5^{2\left|\mathrm{x}\right|}-11·5^{\left|\mathrm{x}\right|}+2=0 \\ \mathrm{Пусть} \mathrm{t}= 5^{\left|\mathrm{x}\right|} \\ 5{\mathrm{t}}^2-11\mathrm{t}+2=0 \\ \mathrm{D} = 121-4\times 2\times 5=121-40 = 81 \\ {\mathrm{t}}_1=\frac{11-9}{10}=\frac{2}{10}=\frac{1}{5} \\ {\mathrm{t}}_2=\frac{11+9}{10}=\frac{20}{10}=2 \\ \mathrm{Решив} \mathrm{неравенство} \mathrm{относительно} \mathrm{t}, \mathrm{получим} \mathrm{совокупность}: \end{gathered}$$

$$\begin{gathered} \left[\begin{array}{c}t<\frac{1}{5}\\ t>2\end{array}\right. \\ Сделаем обратную замену: \\ \left\{\begin{array}{l}\left[\begin{array}{c}{5}^{\left|x\right|}<5^{-1}\\ 5^{\left|x\right|}>5^{{\log }_{5}2}\end{array}\right.\\ x>5\end{array}\right.\iff \left\{\begin{array}{l}\left[\begin{array}{c}\left|x\right|<-1\\ \left|x\right|>{\log }_{5}2\end{array}\right.\\ x>5\end{array}\right.\iff \left\{\begin{array}{l}\left[\begin{array}{c}x\in \varnothing \\ x>-{\log }_{5}2\\ x<-{\log }_{5}2\end{array}\right.\\ x>5\end{array}\right. \\ \iff \left\{\begin{array}{l}x\in (-\infty ;-{\log }_{5}2)\cup (-{\log }_{5}2;+\infty )\\ x>5\end{array}\right.\Rightarrow x\in (5;+\infty ) \end{gathered}$$

Ответ: $(5;+\infty )$