ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 240

\[\boxed{\text{241.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{5a + 7 - 28a^{2}}{20a} = a²\]

\[5a + 7 - 28a^{2} = 20a^{3},\]

\[\ \ a \neq 0\]

\[20a^{3} + 28a^{2} - 5a - 7 = 0\]

\[(5a + 7)\left( 4a^{2} - 1 \right) = 0\]

\[(5a + 7)(2a - 1)(2a + 1) = 0\]

\[a_{1} = - 1,4;\ \ \ \ \ a_{2} = \frac{1}{2},\]

\[\text{\ \ }a_{3} = - \frac{1}{2}.\]

\[Ответ:при\ \ a = - 1,4;\ \]

\[\ a = \pm 0,5.\]

\[\textbf{б)}\ \frac{2 - 18a^{2} - a}{3a} = - 3a²\]

\[2 - 18a² - a = - 9a^{3},\ \ a \neq 0\]

\[9a^{3} - 18a^{2} - a + 2 = 0\]

\[\left( 9a^{2} - 1 \right)(a - 2) = 0\]

\[(3a - 1)(3a + 1)(a - 2) = 0\]

\[a_{1} = \frac{1}{3},\ \ a_{2} = - \frac{1}{3},\ \ \]

\[a_{3} = 2.\]

\[Ответ:при\ \ a = 2;\ \ a = \pm \frac{1}{3}.\]