ext{ Тогда неравенство примет вид:}

$\frac{t}{t-4}-\frac{3}{t-1}-\frac{8}{t^2-5t+4}\ge 0 \iff \frac{t}{t-4}-\frac{3}{t-1}-\frac{8}{(t-4)(t-1)}\ge 0 \iff$ $\iff \frac{t^2-t-3t+12-8}{(t-4)(t-1)}\ge 0 \iff \frac{t^2-4t+4}{(t-4)(t-1)}\ge 0 \iff \frac{(t-2)^2}{(t-4)(t-1)}\ge 0.$

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

$\left[ \begin{aligned} &t<1,\\ &t=2,\\ &t>4 \end{aligned} \right. \iff \left[ \begin{aligned} &8^x<1,\\ &8^x=2,\\ &8^x>4 \end{aligned} \right. \iff \left[ \begin{aligned} &8^x<8^0,\\ &2^{3x}=2^1,\\ &2^{3x}>2^2 \end{aligned} \right. \iff \left[ \begin{aligned} &x<0,\\ &x=\frac13,\\ &x>\frac23. \end{aligned} \right.$

$\text{Таким образом, решением исходного неравенства является: } x\in(-\infty;0)\cup\left\{\frac13\right\}\cup\left(\frac23;\infty\right).$

$\textbf{Ответ: } (-\infty;0)\cup\left\{\frac13\right\}\cup\left(\frac23;\infty\right).$

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Задание №30103 ЕГЭ по Математике