ГДЗ по алгебре 9 класс Макарычев Задание 296

\[\boxed{\text{296}\text{\ (296)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[\ | \cdot 20a;\ \ a \neq 0\]

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

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

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

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

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

\[5a = - 7\ \ \ \ \ \ \ \ \ \ 2a = 1\ \ \ \ \ \ \ \ \ \ \]

\[2a = - 1\]

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

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

\[Ответ:при\ \ a = - 1,4;\ \ a = \pm 0,5.\]

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

\[| \cdot 3a;\ \ a \neq 0\]

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

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

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

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

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

\[3a = 1;\ \ \ \ \ \ \ 3a = - 1;\ \ \ \ \ \ \ \ \ \ \ a = 2\]

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

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