\[\boxed{\text{104\ (104).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \ \frac{2a - 1}{10a^{2} - a - 2}\]
\[10a^{2} - a - 2 = 0\]
\[D = 1 + 4 \cdot 10 \cdot 2 = 81\]
\[a_{1} = \frac{1 + 9}{20} = \frac{1}{2};\ \ a_{2} =\]
\[= \frac{1 - 9}{20} = - \frac{2}{5};\]
\[10a^{2} - a - 2 =\]
\[= 10 \cdot \left( a - \frac{1}{2} \right)\left( a + \frac{2}{5} \right) =\]
\[= (2a - 1)(5a + 2);\]
\[\Longrightarrow \frac{2a - 1}{10a^{2} - a - 2} =\]
\[= \frac{2a - 1}{(2a - 1)(5a + 2)} = \frac{1}{5a + 2}.\]
\[\textbf{б)}\ \frac{6a^{2} - 5a + 1}{1 - 4a^{2}}\]
\[6a^{2} - 5a + 1 = 0\]
\[D = 5^{2} - 4 \cdot 6 \cdot 1 =\]
\[= 25 - 24 = 1\]
\[a_{1} = \frac{5 + 1}{12} = \frac{1}{2};\ \ \ \ a_{2} =\]
\[= \frac{5 - 1}{12} = \frac{1}{3};\]
\[6a^{2} - 5a + 1 =\]
\[= 6 \cdot \left( a - \frac{1}{2} \right)\left( a - \frac{1}{3} \right) =\]
\[= (2a - 1)(3a - 1);\]
\[\Longrightarrow \frac{6a^{2} - 5a + 1}{1 - 4a^{2}} =\]
\[= \frac{(2a - 1)(3a - 1)}{(1 - 2a)(1 + 2a)} = \frac{1 - 3a}{1 + 2a}\text{.\ }\]
\[\boxed{\text{104.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ f(x) = x + \frac{1}{x}\]
\[f( - x) = - x - \frac{1}{x} =\]
\[= - \left( x + \frac{1}{x} \right) = - f(x)\]
\[функция\ нечетная.\]
\[\textbf{б)}\ f(x) = 2x^{3} - x\]
\[f( - x) = 2 \cdot ( - x)^{3} - ( - x) =\]
\[= - 2x^{3} + x = - \left( 2x^{3} - x \right) =\]
\[= - f(x)\]
\[функция\ нечетная.\]