ГДЗ по алгебре 9 класс Макарычев Задание 337

\[\boxed{\text{337\ (337).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{x - 8}{x + 4} > 2\]

\[\frac{x - 8}{x + 4} - 2^{\backslash x + 4} > 0\]

\[\frac{x - 8 - 2x - 8}{x + 4} > 0\]

\[\frac{- x - 16}{x + 4} > 0\]

\[\frac{x + 16}{x + 4} < 0\]

\[(x + 16)(x + 4) < 0\]

\[x \in ( - 16;\ - 4).\]

\[\textbf{б)}\ \frac{3 - x}{x - 2} < 1\]

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

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

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

\[\frac{2x - 5}{x - 2} > 0\]

\[2 \cdot (x - 2)(x - 2,5) > 0\]

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

\[\textbf{в)}\ \frac{7x - 1}{x} > 5\]

\[\frac{7x - 1}{x} - 5^{\backslash x} > 0\]

\[\frac{7x - 1 - 5x}{x} > 0\]

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

\[2 \cdot x \cdot (x - 0,5) > 0\]

\[x \in ( - \infty;0) \cup (0,5; + \infty).\]

\[\textbf{г)}\ \frac{6 - 2x}{x + 4} > 3\]

\[\frac{6 - 2x}{x + 4} - 3^{\backslash x + 4} > 0\]

\[\frac{6 - 2x - 3x - 12}{x + 4} > 0\]

\[\frac{- 5x - 6}{x + 4} > 0\]

\[\frac{5x + 6}{x + 4} < 0\]

\[5 \cdot (x + 1,2)(x + 4) < 0\]

\[x \in ( - 4; - 1,2).\]