Решебник по алгебре 9 класс Мерзляк | Страница 133

\[\boxed{\text{Упражнения}\text{\ }\text{стр}\text{.\ 133.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\boxed{\text{1.\ }}\]

\[1)\ (x + 1)(x - 2)(x + 5) > 0\]

\[x = - 1;\ \ x = 2;\ \ x = - 5\]

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

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

\[x = 0;\ \ x = 3;\ \ x = - 2\]

\[Ответ:x \in ( - \infty;\ - 2) \in (0;3).\]

\[3)\ (2x - 1)(3 - x)(x + 1) < 0\]

\[x = 0,5;\ \ \ x = 3;\ \ x = - 1\]

\[Ответ:x \in ( - 1;0,5) \cup (3; + \infty).\]

\[4)\ (2x + 3)(3x - 1)(x + 4) > 0\]

\[x = - 1,5;\ \ \ x = \frac{1}{3};\ \ x = - 4\]

\[Ответ:x \in ( - 4;\ - 1,5) \cup\]

\[\cup \left( \frac{1}{3}; + \infty \right).\]

\[\boxed{\text{2.\ }}\]

\[1)\ (2x + 1)(x - 3)\left( x^{2} + 4 \right) < 0\]

\[x = - 0,5;\ \ \ x = 3;\ \ x^{2} \neq - 4\]

\[Ответ:x \in ( - 0,5;3).\]

\[2)\ (2 - x)(3x + 5)\left( x^{2} - x + 1 \right) > 0\]

\[x = 2;\ \ \ x = - \frac{5}{3} = - 1\frac{2}{3}\]

\[x^{2} - x + 1 = 0\]

\[D = 1 - 4 < 0 \Longrightarrow нет\ корней.\]

\[Ответ:x \in \left( - 1\frac{2}{3};3 \right).\]

\[3)\ (2x + 1)^{2}\left( x^{2} - 4x + 3 \right) > 0\]

\[(2x + 1)^{2} > 0 \Longrightarrow при\ любом\ x;\]

\[2x = - 1\]

\[x = - 0,5\]

\[x^{2} - 4x + 3 = 0\]

\[x_{1} = 1;\ \ x = 3\]

\[Ответ:x \in ( - \infty; - 0,5) \cup\]

\[\cup ( - 0,5;1) \cup (3; + \infty).\]

\[\boxed{\text{3.\ }}\]

\[1)\ \frac{x + 3}{x - 1} > 0\]

\[x = - 3;\ \ \ x \neq 1\]

\[Ответ:x \in ( - \infty;\ - 3) \cup (1;\ + \infty).\]

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

\[x = 2;\ \ x = - 1;\ \ \ x \neq 4\]

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

\[3)\ \frac{(2x + 1)(x - 3)}{(2 - x)(x - 5)} < 0\]

\[x = - 0,5;\ \ \ \ x = 3;\ \ \ \ \]

\[x \neq 2;\ \ \ \ x \neq 5\]

\[Ответ:x( - \infty;\ - 0,5) \cup (2;3) \cup\]

\[\cup (5; + \infty).\]

\[4)\ \frac{x^{3}(x - 1)^{4}(x + 5)}{(x - 8)(1 - 4x)} > 0\]

\[(x - 1)^{4} > 0 \Longrightarrow при\ любом\ x;\]

\[x = 0;\ \ \ x = - 5;\]

\[x \neq 8;\ \ x \neq 0,25\]

\[Ответ:x \in ( - 5;0) \cup (0,25;8).\]

\[\boxed{\text{4.\ }}\]

\[1)\ \frac{1}{x} < 1\]

\[1 < x;\ \ \ \ x \neq 0\]

\[x > 1\]

\[Ответ:x\nexists(1;\ + \infty).\]

\[2)\ \frac{x}{x + 3} > \frac{1}{2}\text{\ \ \ \ \ \ }\]

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

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

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

\[x = 3;\ \ \ x \neq - 3\]

\[Ответ:x \in ( - \infty;\ - 3) \cup (3; + \infty).\]

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

\[\frac{x - 3 - 3x - 6}{(x + 2)(x - 3)} < 0\]

\[\frac{- 2x - 9}{(x + 2)(x - 3)} < 0\]

\[- 2x - 9 = 0\]

\[x = - 4,5\]

\[x \neq - 2;\ \ \ x \neq 3\]

\[Ответ:x \in ( - 4,5;\ - 2) \cup\]

\[\cup (3;\ + \infty).\]

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

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

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

\[x^{2} - 2x + 5 > 0\ при\ любом\ x,\]

\[\ так\ как\ a = 1 > 0;\ \ D < 0.\]

\[1 - x^{2} \neq 0\]

\[x \neq 1;\ \ \ x \neq - 1\]

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