ГДЗ по алгебре 9 класс Макарычев Задание 104

\[\boxed{\text{104\ (104).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \ \frac{2a - 1}{10a^{2} - a - 2}\]

\[10a^{2} - a - 2 = 0\]

\[D = 1 + 4 \cdot 10 \cdot 2 = 81\]

\[a_{1} = \frac{1 + 9}{20} = \frac{1}{2};\ \ a_{2} =\]

\[= \frac{1 - 9}{20} = - \frac{2}{5};\]

\[10a^{2} - a - 2 =\]

\[= 10 \cdot \left( a - \frac{1}{2} \right)\left( a + \frac{2}{5} \right) =\]

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

\[\Longrightarrow \frac{2a - 1}{10a^{2} - a - 2} =\]

\[= \frac{2a - 1}{(2a - 1)(5a + 2)} = \frac{1}{5a + 2}.\]

\[\textbf{б)}\ \frac{6a^{2} - 5a + 1}{1 - 4a^{2}}\]

\[6a^{2} - 5a + 1 = 0\]

\[D = 5^{2} - 4 \cdot 6 \cdot 1 =\]

\[= 25 - 24 = 1\]

\[a_{1} = \frac{5 + 1}{12} = \frac{1}{2};\ \ \ \ a_{2} =\]

\[= \frac{5 - 1}{12} = \frac{1}{3};\]

\[6a^{2} - 5a + 1 =\]

\[= 6 \cdot \left( a - \frac{1}{2} \right)\left( a - \frac{1}{3} \right) =\]

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

\[\Longrightarrow \frac{6a^{2} - 5a + 1}{1 - 4a^{2}} =\]

\[= \frac{(2a - 1)(3a - 1)}{(1 - 2a)(1 + 2a)} = \frac{1 - 3a}{1 + 2a}\text{.\ }\]