Тип 15 № 15723

Решите неравенство $3\cdot25^{1-\frac{3}{x}} - 152\cdot15^{-\frac{3}{x}} + 5\cdot3^{2 - \frac{6}{x}} > 0$

Источник: Сборник И.В. Ященко 2026. Вариант 12.

Решение:

$(*) \quad x \neq 0$ 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: $(-3; 0) \cup (0; 1,5)$