б) $$ \frac{6a+1}{(2a+1)(2a-1)} - \frac{2a}{-(2a+1)} = \frac{6a+1}{(2a+1)(2a-1)} + \frac{2a}{2a+1} = \frac{6a+1 + 2a(2a-1)}{(2a+1)(2a-1)} = \frac{6a+1 + 4a^2 - 2a}{(2a+1)(2a-1)} = \frac{4a^2 + 4a + 1}{(2a+1)(2a-1)} = \frac{(2a+1)^2}{(2a+1)(2a-1)} = \frac{2a+1}{2a-1} $$
Ответ: $$ \frac{2a+1}{2a-1} $$