Тип 15 № 11225

$$2^x + \frac{4\cdot2^x}{2^x -4} + \frac{2^{2x} + 7\cdot2^x + 20}{2^{2x} - 12\cdot2^x + 32} \leq 1$$

Пусть $2^x = t > 0$

Имеем: $$t + \frac{4t}{t -4} + \frac{t^2 + 7t + 20}{t^2 - 12t + 32} \leq 1$$ $$t + \frac{4t(t - 8)}{t -4} + \frac{t^2 + 7t + 20}{(t - 4)(t - 8)} - 1 \leq 0$$ $$t + \frac{4t(t - 8)}{(t -4)(t - 8)} + \frac{t^2 + 7t + 20 - t^2 + 12t - 32}{(t - 4)(t - 8)}\leq 0$$ $$t + \frac{4t(t - 8)}{(t -4)(t - 8)} + \frac{19t - 12}{(t - 4)(t - 8)}\leq 0$$ $$t + \frac{4t(t - 8)}{(t -4)(t - 8)} + \frac{19t - 12}{(t - 4)(t - 8)}\leq 0$$ $$t + \frac{4t^2 -32t + 19t - 12}{(t -4)(t - 8)} \leq 0$$ $$t + \frac{4t^2 -13t - 12}{(t -4)(t - 8)} \leq 0$$ $$\frac{t(t - 4)(t - 8) + (t - 4)(4t + 3)}{(t -4)(t - 8)} \leq 0$$ $$\frac{(t - 4)(t^2 - 8t + 4t + 3)}{(t -4)(t - 8)} \leq 0$$ $$\frac{(t - 4)(t - 1)(t - 3)}{(t -4)(t - 8)} \leq 0$$

Решение данного неравенства равносильно совокупности:

$$\left[\begin{aligned} &\begin{cases} t \geq 3\\ t < 4 \end{cases}\\ &\begin{cases} t > 4\\ t < 8 \end{cases}\\ &t \leq 1 \\ \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &\begin{cases} 2^x \geq 3\\ 2^x < 4 \end{cases}\\ &\begin{cases} 2^x > 4\\ 2^x < 8 \end{cases}\\ &2^x \leq 1 \\ \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &\begin{cases} x \geq \log_23\\ x < 2 \end{cases}\\ &\begin{cases} x > 2\\ x < 3 \end{cases}\\ &x \leq 0 \\ \end{aligned}\right.$$

Ответ: $(-\infty; 0] \cup [\log_23; 2) \cup (2; 3)$