Выполните умножение:
1) $\frac{3a + b}{4c} * \frac{c}{3a + b}$;
2) $\frac{ab - b^2}{8} * \frac{4a}{b^4}$;
3) $\frac{5x - 5y}{x^6} * \frac{x^3}{x - y}$;
4) $\frac{18b}{b^2 - 16} * \frac{b + 4}{3b}$;
5) $\frac{6}{m^2 - 9n^2} * (m - 3n)$;
6) $\frac{3c - 9}{9c^2 + 6c + 1} * \frac{3c + 1}{c - 3}$.
$\frac{3a + b}{4c} * \frac{c}{3a + b} = \frac{1}{4} * \frac{1}{1} = \frac{1}{4}$
$\frac{ab - b^2}{8} * \frac{4a}{b^4} = \frac{b(a - b)}{8} * \frac{4a}{b^4} = \frac{a - b}{2} * \frac{a}{b^3} = \frac{a(a - b)}{2b^3}$
$\frac{5x - 5y}{x^6} * \frac{x^3}{x - y} = \frac{5(x - y)}{x^6} * \frac{x^3}{x - y} = \frac{5}{x^3} * \frac{1}{1} = \frac{5}{x^3}$
$\frac{18b}{b^2 - 16} * \frac{b + 4}{3b} = \frac{18b}{(b - 4)(b + 4)} * \frac{b + 4}{3b} = \frac{6}{b - 4} * \frac{1}{1} = \frac{6}{b - 4}$
$\frac{6}{m^2 - 9n^2} * (m - 3n) = \frac{6}{(m - 3n)(m + 3n)} * (m - 3n) = \frac{6}{m + 3n} * 1 = \frac{6}{m + 3n}$
$\frac{3c - 9}{9c^2 + 6c + 1} * \frac{3c + 1}{c - 3} = \frac{3(c - 3)}{(3c + 1)^2} * \frac{3c + 1}{c - 3} = \frac{3}{3c + 1} * \frac{1}{1} = \frac{3}{3c + 1}$
Пожауйста, оцените решение