Выполните действия:
а) $\frac{2b}{2b + 3} - \frac{5}{3 - 2b} - \frac{4b^2 + 9}{4b^2 - 9}$;
б) $\frac{c + 6b}{ac + 2bc - 6ab - 3a^2} + \frac{2b}{a^2 + 2ab} - \frac{b}{ac - 3a^2}$.
$\frac{2b}{2b + 3} - \frac{5}{3 - 2b} - \frac{4b^2 + 9}{4b^2 - 9} = \frac{2b}{2b + 3} + \frac{5}{2b - 3} - \frac{4b^2 + 9}{(2b - 3)(2b + 3)} = \frac{2b(2b - 3) + 5(2b + 3) - 4b^2 - 9}{(2b - 3)(2b + 3)} = \frac{4b^2 - 6b + 10b + 15 - 4b^2 - 9}{(2b - 3)(2b + 3)} = \frac{4b + 6}{(2b - 3)(2b + 3)} = \frac{2(2b + 3)}{(2b - 3)(2b + 3)} = \frac{2}{2b - 3}$
$\frac{c + 6b}{ac + 2bc - 6ab - 3a^2} + \frac{2b}{a^2 + 2ab} - \frac{b}{ac - 3a^2} = \frac{c + 6b}{c(a + 2b) - 3a(2b + a)} + \frac{2b}{a(a + 2b)} - \frac{b}{a(c - 3a)} = \frac{c + 6b}{(a + 2b)(c - 3a)} + \frac{2b}{a(a + 2b)} - \frac{b}{a(c - 3a)} = \frac{a(c + 6b) + 2b(c - 3a) - b(a + 2b)}{a(a + 2b)(c - 3a)} = \frac{ac + 6ab + 2bc - 6ab - ab - 2b^2}{a(a + 2b)(c - 3a)} = \frac{ac + 2bc - ab - 2b^2}{a(a + 2b)(c - 3a)} = \frac{c(a + 2b) - b(a + 2b)}{a(a + 2b)(c - 3a)} = \frac{(a + 2b)(c - b)}{a(a + 2b)(c - 3a)} = \frac{c - b}{a(c - 3a)}$
Пожауйста, оцените решение
