display">\frac{t}{t-4}+\frac{t+5}{t-5}+\frac{22}{t^2-9t+20}\le 0 \iff \frac{t}{t-4}+\frac{t+5}{t-5}+\frac{22}{(t-4)(t-5)}\le 0 \iff $\iff \frac{t(t-5)+(t+5)(t-4)+22}{(t-4)(t-5)}\le 0 \iff \frac{t^2-5t+t^2-4t+5t-20+22}{(t-4)(t-5)}\le 0 \iff$ $\iff \frac{2t^2-4t+2}{(t-4)(t-5)}\le 0 \iff \frac{2(t-1)^2}{(t-4)(t-5)}\le 0.$

$\text{Решим полученное неравенство методом интервалов:}$

$\left[ \begin{aligned} &t=1,\\ &4<t<5 \end{aligned} \right. \iff \left[ \begin{aligned} &5^x=1,\\ &4<5^x<5 \end{aligned} \right. \iff \left[ \begin{aligned} &5^x=5^0,\\ &5^{\log_5 4}<5^x<5^1 \end{aligned} \right. \iff \left[ \begin{aligned} &x=0,\\ &\log_5 4<x<1. \end{aligned} \right.$

$\text{Таким образом, решением исходного неравенства является: } x\in\{0\}\cup(\log_5 4;1).$

$\textbf{Ответ: } \{0\}\cup(\log_5 4;1).$

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Задание №30102 ЕГЭ по Математике