ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1391

\[\boxed{\mathbf{1391}\mathbf{.}}\]

\[1)\ \frac{5x + 4}{x - 3} < 4\]

\[\frac{5x + 4 - 4(x - 3)}{x - 3} < 0\]

\[\frac{5x + 4 - 4x + 12}{x - 3} < 0\]

\[\frac{x + 16}{x - 3} < 0\]

\[(x + 16)(x - 3) < 0\]

\[- 16 < x < 3.\]

\[Ответ:\ \ x \in ( - 16;\ 3).\]

\[2)\ \frac{2}{x - 4} < 1\]

\[\frac{2 - (x - 4)}{x - 4} < 0\]

\[\frac{2 - x + 4}{x - 4} < 0\]

\[\frac{6 - x}{x - 4} < 0\]

\[(x - 4)(6 - x) < 0\]

\[(x - 4)(x - 6) > 0\]

\[x < 4\ \ и\ \ x > 6.\]

\[Ответ:\ \ x \in ( - \infty;\ 4) \cup (6;\ + \infty).\]

\[3)\ \frac{2}{x + 3} \leq 4\]

\[\frac{2 - 4(x + 3)}{x + 3} \leq 0\]

\[\frac{2 - 4x - 12}{x + 3} \leq 0\]

\[\frac{4x + 10}{x + 3} \geq 0\]

\[(x + 3)(4x + 10) \geq 0\]

\[x < - 3\ \ и\ \ x \geq - 2,5.\]

\[Ответ:\ \ \]

\[x \in ( - \infty;\ - 3) \cup \lbrack - 2,5;\ + \infty).\]