Запишите в виде дробей с одинаковыми знаменателями:
1) $\frac{3a}{3a - 2}, \frac{a}{9a + 6}$ и $\frac{a^2}{9a^2b - 4b}$;
2) $\frac{1}{a - 5b}, \frac{1}{a^2 + 7ac}$ и $\frac{1}{a^2 + 7ac - 5ab - 35bc}$.
$\frac{3a}{3a - 2} = \frac{3a * 3b(3a + 2)}{3b(3a - 2)(3a + 2)} = \frac{9ab(3a + 2)}{3b(9a^2 - 4)}$
$\frac{a}{9a + 6} = \frac{a}{3(3a + 2)} = \frac{a * b(3a - 2)}{3b(3a - 2)(3a + 2)} = \frac{ab(3a - 2)}{3b(9a^2 - 4)}$
$\frac{a^2}{9a^2b - 4b} = \frac{a^2}{b(9a^2 - 4)} = \frac{a^2}{b(3a - 2)(3a + 2)} = \frac{3a^2}{3b(9a^2b - 4b)}$
$\frac{1}{a - 5b} = \frac{a(a + 7c)}{a(a - 5b)(a + 7c)}$
$\frac{1}{a^2 + 7ac} = \frac{1}{a(a + 7c)} = \frac{a - 5b}{a(a - 5b)(a + 7c)}$
$\frac{1}{a^2 + 7ac - 5ab - 35bc} = \frac{1}{(a^2 + 7ac) - (5ab + 35bc)} = \frac{1}{a(a + 7c) - 5b(a + 7c)} = \frac{a}{a(a + 7c)(a - 5b)}$
Пожауйста, оцените решение
