Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1610

\[\boxed{\mathbf{1610}\mathbf{.}}\]

\[1)\ \frac{2x - 3}{4 - x} > \frac{1}{x}\]

\[\frac{2x - 3}{4 - x} - \frac{1}{x} > 0\]

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

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

\[2 \bullet \frac{x^{2} - x - 2}{x(4 - x)} > 0\]

\[\frac{x^{2} + x - 2x - 2}{x(4 - x)} > 0\]

\[\frac{x(x + 1) - 2(x + 1)}{x(4 - x)} > 0\]

\[\frac{(x - 2)(x + 1)}{x(x - 4)} < 0\]

\[(x + 1) \bullet x \bullet (x - 2) \bullet (x - 4) < 0\]

\[- 1 < x < 0\ \ и\ \ 2 < x < 4.\]

\[Ответ:\ \ x \in ( - 1;\ 0) \cup (2;\ 4).\]

\[2)\ \frac{2x + 5}{|x + 1|} \geq 1\]

\[x + 1 > 0\]

\[x > - 1.\]

\[При\ x > - 1:\]

\[\frac{2x + 5}{x + 1} \geq 1\]

\[\frac{2x + 5}{x + 1} - 1 \geq 0\]

\[\frac{2x + 5 - (x + 1)}{x + 1} \geq 0\]

\[\frac{x + 4}{x + 1} \geq 0\]

\[(x + 4)(x + 1) \geq 0\]

\[x \leq - 4\ \ и\ \ x > - 1.\]

\[При\ x < - 1:\]

\[\frac{2x + 5}{- (x + 1)} \geq 1\]

\[\frac{2x + 5}{x + 1} \leq - 1\]

\[\frac{2x + 5}{x + 1} + 1 \leq 0\]

\[\frac{2x + 5 + x + 1}{x + 1} \leq 0\]

\[3 \bullet \frac{x + 2}{x + 1} \leq 0\]

\[(x + 2)(x + 1) \leq 0\]

\[- 2 \leq x < - 1.\]

\[Ответ:\ \ \]

\[x \in \lbrack - 2;\ - 1) \cup ( - 1;\ + \infty).\]