ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 131

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

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

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

\[a_{1,2} = \frac{1 \pm 9}{20},\ \ a_{1} = \frac{1}{2},\]

\[\text{\ \ }a_{2} = - \frac{2}{5};\]

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

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

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

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

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

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

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

\[D = 5^{2} - 4 \cdot 6 \cdot 1 = 25 - 24 = 1\]

\[a_{1,2} = \frac{5 \pm 1}{12},\ \ a_{1} = \frac{1}{2},\ \ \]

\[a_{2} = \frac{1}{3};\]

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

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

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

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

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