Показательные неравенства заменой

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

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

Сделаем замену $t = 2^{6x - x^2 - 4} > 0$. Тогда неравенство примет вид: $$t^2 - 34t + 64 \geq 0 \Leftrightarrow(t - 2)(t - 32) \geq 0$$

Решение неравенства: $$t \leq 2 \quad \text{или} \quad t \geq 32$$

1. $2^{6x - x^2 - 4} \leq 2^1$ $$6x - x^2 - 4 \leq 1 \Rightarrow x^2 - 6x + 5 \geq 0 \Rightarrow x \in (-\infty;1] \cup [5;+\infty)$$

2. $2^{6x - x^2 - 4} \geq 2^5$ $$6x - x^2 - 4 \geq 5 \Leftrightarrow x^2 - 6x + 9 \Leftrightarrow (x - 3)^2 \leq 0 \Rightarrow x = 3$$

Ответ: $x \in (-\infty;1] \cup \{3\} \cup [5;+\infty)$

Пусть $5^x - 10 = t > 0$

$$t \geq \frac{225}{t} \Leftrightarrow \frac{t^2 - 225}{t} \geq 0 \Leftrightarrow \frac{(t - 15)(t + 15)}{t} \geq 0$$

$$\left[\begin{aligned} &\begin{cases} t \geq -15 \\ t < 0 \end{cases}\\ &t \geq 15 \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &\begin{cases} 5^x - 10 \geq -15 \\ 5^x - 10 < 0 \end{cases}\\ &5^x - 10 \geq 15 \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &5^x < 10\\ &5^x \geq 5^2 \end{aligned}\right. \Leftrightarrow$$ $$\Leftrightarrow \left[\begin{aligned} &5^x < 5^{\log_510}\\ &5^x \geq 5^2 \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &x < \log_510\\ &x \geq 2 \end{aligned}\right. \Rightarrow x \in (-\infty; \log_510) \cup [2; +\infty)$$

Ответ: $(-\infty; \log_510) \cup [2; +\infty)$

$$(*) x \neq 0$$ Преобразуем неравенство: $$80\cdot5^{-\frac{8}{x}} - 189\cdot5^{-\frac{4}{x}}\cdot4^{-\frac{4}{x}} + 100\cdot4^{-\frac{8}{x}} \leq 0$$ Домножим неравенство на $5^{\frac{4}{x}}\cdot4^{\frac{4}{x}} > 0$: $$80\cdot5^{-\frac{4}{x}}\cdot4^{\frac{4}{x}} - 189 + 100\cdot4^{-\frac{4}{x}}\cdot5^{\frac{4}{x}} \leq 0$$ $$80\cdot\left(\frac{4}{5}\right)^{\frac{4}{x}} - 189 + 100 \cdot\left(\frac{5}{4}\right)^{\frac{4}{x}} \leq 0$$ Пусть $t = \left(\frac{5}{4}\right)^{\frac{4}{x}} > 0$

Тогда: $$100t + \frac{80}{t} - 189 \leq 0 \Leftrightarrow 100t^2 -189t + 80 \leq 0$$ $$D = 189^2 - 4\cdot100\cdot80 = 35721 - 32000 =3721 = 61^2$$ $$t_1 = \frac{189 + 61}{200} = \frac{5}{4}$$ $$t_2 = \frac{189 - 61}{200} = \frac{128}{200} = \frac{16}{25}$$ $$\left(t - \frac{16}{25}\right)\left(t - \frac{5}{4}\right) \leq 0 \Rightarrow t \in \left[\frac{16}{25}; \frac{5}{4}\right]$$ $$\begin{cases} t \geq \frac{16}{25}\\ t \leq \frac{5}{4} \end{cases} \Leftrightarrow \begin{cases} \left(\frac{5}{4}\right)^{\frac{4}{x}} \geq \frac{16}{25}\\ \left(\frac{5}{4}\right)^{\frac{4}{x}} \leq \frac{5}{4} \end{cases} \Leftrightarrow \begin{cases} \left(\frac{5}{4}\right)^{\frac{4}{x}} \geq \left(\frac{5}{4}\right)^{-2}\\ \left(\frac{5}{4}\right)^{\frac{4}{x}} \leq \left(\frac{5}{4}\right)^1 \end{cases} \Leftrightarrow$$ $$\Leftrightarrow \begin{cases} \tfrac{4}{x} \geq -2\\ \tfrac{4}{x} \leq 1 \end{cases} \Leftrightarrow \begin{cases} \frac{x + 2}{x} \geq 0\\ \frac{x - 4}{x} \geq 0 \end{cases} \Leftrightarrow \begin{cases} x \in (-\infty; -2] \cup (0; +\infty)\\ x \in (-\infty; 0) \cup [4; +\infty) \end{cases}$$ Пересекая решения, получаем ответ: $$x \in (-\infty; -2] \cup [4; +\infty)$$

Ответ: $(-\infty; -2] \cup [4; +\infty)$

$$(*) \quad x \neq 0$$ Преобразуем неравенство: $$3\cdot25\cdot25^{-\frac{3}{x}} - 152\cdot5^{-\frac{3}{x}}\cdot3^{-\frac{3}{x}} + 5\cdot9\cdot3^{-\frac{6}{x}} > 0$$ $$75\cdot25^{-\frac{3}{x}} - 152\cdot5^{-\frac{3}{x}}\cdot3^{-\frac{3}{x}} + 45\cdot3^{-\frac{6}{x}} > 0$$ Домножим неравенство на $5^{\frac{3}{x}}\cdot3^{\frac{3}{x}} > 0$: $$75\cdot25^{-\frac{3}{x}}\cdot5^{\frac{3}{x}}\cdot3^{\frac{3}{x}} - 152 + 45\cdot3^{-\frac{6}{x}}\cdot5^{\frac{3}{x}}\cdot3^{\frac{3}{x}} > 0$$ $$75\cdot(5^{-2})^{\frac{3}{x}}\cdot5^{\frac{3}{x}}\cdot3^{\frac{3}{x}} - 152 + 45\cdot3^{-\frac{6}{x}}\cdot5^{\frac{3}{x}}\cdot3^{\frac{3}{x}} > 0$$ $$75\cdot5^{-\frac{6}{x}}\cdot5^{\frac{3}{x}}\cdot3^{\frac{3}{x}} - 152 + 45\cdot3^{-\frac{3}{x}}\cdot5^{\frac{3}{x}} > 0$$ $$75\cdot5^{-\frac{3}{x}}\cdot3^{\frac{3}{x}} - 152 + 45\cdot3^{-\frac{3}{x}}\cdot5^{\frac{3}{x}} > 0$$ $$75\cdot\left(\frac{3}{5}\right)^{\frac{3}{x}} - 152 + 45\cdot\left(\frac{5}{3}\right)^{\frac{3}{x}} > 0$$ Пусть $t = \left(\frac{5}{3}\right)^{\frac{3}{x}} > 0$. Тогда $\left(\frac{3}{5}\right)^{\frac{3}{x}} = \frac{1}{t}$.

$$75\cdot\frac{1}{t} - 152 + 45t > 0 \quad \Big| \cdot t > 0$$ $$45t^2 - 152t + 75 > 0$$ $$D = 152^2 - 4\cdot45\cdot75 = 23104 - 13500 = 9604 = 98^2$$ $$t_{1,2} = \frac{152 \pm 98}{2\cdot45} = \frac{152 \pm 98}{90}$$ $$t_1 = \frac{152 - 98}{90} = \frac{54}{90} = \frac{3}{5}, \quad t_2 = \frac{152 + 98}{90} = \frac{250}{90} = \frac{25}{9}$$ $$(t - \tfrac{3}{5})(t - \tfrac{25}{9}) > 0$$ $$t \in (-\infty; \tfrac{3}{5}) \cup (\tfrac{25}{9}; +\infty)$$

$$\left[\begin{aligned} &\left(\frac{5}{3}\right)^{\frac{3}{x}} < \left(\frac{3}{5}\right)\\ &\left(\frac{5}{3}\right)^{\frac{3}{x}} > \left(\frac{5}{3}\right)^2 \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &\left(\frac{5}{3}\right)^{\frac{3}{x}} < \left(\frac{5}{3}\right)^{-1}\\ &\left(\frac{5}{3}\right)^{\frac{3}{x}} > \left(\frac{5}{3}\right)^2 \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} &\tfrac{3}{x} < -1\\ &\tfrac{3}{x} > 2 \end{aligned}\right. \Leftrightarrow$$ $$\Leftrightarrow \left[\begin{aligned} \tfrac{x + 3}{x} < 0\\ \tfrac{3 - 2x}{x} >0 \end{aligned}\right. \Leftrightarrow \left[\begin{aligned} x \in (-3; 0) \\ x \in \left(0; 1,5\right) \end{aligned}\right.$$ Объединяя решения, получаем ответ: $$x \in (-3; 0) \cup (0; 1,5)$$

Ответ: $(-3; 0) \cup (0; 1,5)$