Выполните действия:
а) $\frac{a^2 - 9}{2a^2 + 1} * (\frac{6a + 1}{a - 3})$;
б) $(\frac{5x + y}{x - 5y} + \frac{5x - y}{x + 5y}) : \frac{x^2 + y^2}{x^2 - 25y^2}$.
$\frac{a^2 - 9}{2a^2 + 1} * (\frac{6a + 1}{a - 3}) = \frac{(a - 3)(a + 3)}{2a^2 + 1} * \frac{(6a + 1)(a + 3) + (6a - 1)(a - 3)}{(a - 3)(a + 3)} = \frac{(6a + 1)(a + 3) + (6a - 1)(a - 3)}{2a^2 + 1} = \frac{6a^2 + 19a + 3 + 6a^2 - 19a + 3}{2a^2 + 1} = \frac{12a^2 + 6}{2a^2 + 1} = \frac{6(2a^2 + 1)}{2a^ + 1} = 6$
$(\frac{5x + y}{x - 5y} + \frac{5x - y}{x + 5y}) : \frac{x^2 + y^2}{x^2 - 25y^2} = \frac{(5x + y)(x + 5y) + (5x - y)(x - 5y)}{(x - 5y)(x + 5y)} * \frac{(x - 5y)(x + 5y)}{x^2 + y^2} = \frac{5x^2 + 26xy + 5y^2 + 5x^2 - 26xy + 5y^2}{x^2 + y^2} = \frac{10(x^2 + y^2)}{x^2 + y^2} = 10$
Пожауйста, оцените решение
