ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 297

\[\boxed{\text{298.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{5x + 4}{x} < 4^{\backslash x}\]

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

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

\[x(x + 4) < 0\]

\[x \in ( - 4;0).\]

\[\textbf{б)}\ \frac{6x + 1}{x + 1} > 1^{\backslash x + 1}\]

\[\frac{6x + 1 - x - 1}{x + 1} > 0\]

\[\frac{5x}{x + 1} > 0\]

\[5x(x + 1) > 0\]

\[x \in ( - \infty;\ - 1) \cup (0; + \infty).\]

\[\textbf{в)}\ \frac{x}{x - 1} \geq 2^{\backslash x - 1}\]

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

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

\[\frac{x - 2}{x - 1} \leq 0\]

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

\[x \in (1;2\rbrack.\]

\[\textbf{г)}\ \frac{3x - 1}{x + 2} \geq 1\]

\[\frac{3x - 1}{x + 2} - 1^{\backslash x + 2} \geq 0\]

\[\frac{3x - 1 - x - 2}{x + 2} \geq 0\]

\[\frac{2x - 3}{x + 2} \geq 0\]

\[(2x - 3)(x + 2) \geq 0\]

\[2 \cdot (x + 2)(x - 1,5) \geq 0\]

\[x \in ( - \infty;\ - 2) \cup \lbrack 1,5;\ + \infty).\]