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МатематикаРешебник по алгебре 9 класс Макарычев Задание 296
Задание 296
\[\boxed{\text{296}\text{\ (296)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{5a + 7 - 28a^{2}}{20a} = a^{2}\text{\ \ \ \ \ \ \ }\]
\[\ | \cdot 20a;\ \ a \neq 0\]
\[5a + 7 - 28a^{2} = 20a^{3}\ \]
\[20a^{3} + 28a^{2} - 5a - 7 = 0\]
\[4a^{2}(5a + 7) - (5a + 7) = 0\]
\[(5a + 7)\left( 4a^{2} - 1 \right) = 0\]
\[(5a + 7)(2a - 1)(2a + 1) = 0\]
\[5a = - 7\ \ \ \ \ \ \ \ \ \ 2a = 1\ \ \ \ \ \ \ \ \ \ \]
\[2a = - 1\]
\[a_{1} = - 1,4\text{\ \ \ \ \ \ \ }a_{2} = \frac{1}{2}\text{\ \ \ \ \ \ }\]
\[a_{3} = - \frac{1}{2}.\]
\[Ответ:при\ \ a = - 1,4;\ \ a = \pm 0,5.\]
\[\textbf{б)}\ \frac{2 - 18a^{2} - a}{3a} = - 3a^{2}\text{\ \ \ \ \ \ }\]
\[| \cdot 3a;\ \ a \neq 0\]
\[2 - 18a^{2} - a = - 9a^{3}\]
\[9a^{3} - 18a^{2} - a + 2 = 0\]
\[9a^{2}(a - 2) - (a - 2) = 0\]
\[\left( 9a^{2} - 1 \right)(a - 2) = 0\]
\[(3a - 1)(3a + 1)(a - 2) = 0\]
\[3a = 1;\ \ \ \ \ \ \ 3a = - 1;\ \ \ \ \ \ \ \ \ \ \ a = 2\]
\[a_{1} = \frac{1}{3}\text{\ \ \ \ \ \ \ \ }a_{2} = - \frac{1}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ }a_{3} = 2.\]
\[Ответ:при\ \ a = 2;\ \ a = \pm \frac{1}{3}.\]
\[\boxed{\text{296.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \frac{x - 1}{x - 3} \geq 0\]
\[\left\{ \begin{matrix} (x - 1)(x - 3) \geq 0 \\ x \neq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\ x \in ( - \infty;1\rbrack \cup (3; + \infty).\]
\[\textbf{б)}\ \frac{x + 6}{x - 5} \leq 0\]
\[\left\{ \begin{matrix} (x + 6)(x - 5) \leq 0 \\ x \neq 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ }\]
\[x \in \lbrack - 6;5).\]
\[\textbf{в)}\ \frac{2 - x}{x} \geq 0\]
\[\left\{ \begin{matrix} x(2 - x) \geq 0 \\ x \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x(x - 2) \leq 0 \\ x \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ }\]
\[x \in (0;2\rbrack.\]
\[\textbf{г)}\ \frac{3 - 2x}{x - 1} \leq 0\]
\[\left\{ \begin{matrix} (3 - 2x)(x - 1) \leq 0 \\ x \neq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \ \left\{ \begin{matrix} 2 \cdot (x - 1)(x - 1,5) \geq 0 \\ x \neq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x \in ( - \infty;1) \cup \lbrack 1,5; + \infty).\]
