Преобразуйте в алгебраическую дробь:
а) $\frac{a^2 - b^2}{2a^2b} * \frac{4ab^2}{a + b}$;
б) $\frac{(x - y)^2}{3x^2y^3} : \frac{x - y}{6xy^2}$;
в) $\frac{mn - m^2}{2m} * \frac{8n}{n^2 - m^2}$;
г) $\frac{2a - 4}{b + 1} : \frac{a^2 - 4}{(b + 1)^2}$;
д) $\frac{x + y}{x - y} * \frac{x^2 - xy}{2x^2 - 2y^2}$;
е) $\frac{16 - m^2}{m^2 - 3m} : \frac{m^2 + 4m}{m^2 - 9}$.
$\frac{a^2 - b^2}{2a^2b} * \frac{4ab^2}{a + b} = \frac{(a - b)(a + b)}{2a^2b} * \frac{4ab^2}{a + b} = \frac{a - b}{a} * \frac{2b}{1} = \frac{2b(a - b)}{a}$
$\frac{(x - y)^2}{3x^2y^3} : \frac{x - y}{6xy^2} = \frac{(x - y)^2}{3x^2y^3} * \frac{6xy^2}{x - y} = \frac{x - y}{xy} * \frac{2}{1} = \frac{2(x - y)}{xy}$
$\frac{mn - m^2}{2m} * \frac{8n}{n^2 - m^2} = \frac{m(n - m)}{2m} * \frac{8n}{(n - m)(n + m)} = \frac{n - m}{2} * \frac{8n}{(n - m)(n + m)} = \frac{1}{1} * \frac{4n}{n + m} = \frac{4n}{n + m}$
$\frac{2a - 4}{b + 1} : \frac{a^2 - 4}{(b + 1)^2} = \frac{2a - 4}{b + 1} * \frac{(b + 1)^2}{a^2 - 4} = \frac{2(a - 2)}{b + 1} * \frac{(b + 1)^2}{(a - 2)(a + 2)} = \frac{2}{1} * \frac{b + 1}{a + 2} = \frac{2(b + 1)}{a + 2}$
$\frac{x + y}{x - y} * \frac{x^2 - xy}{2x^2 - 2y^2} = \frac{x + y}{x - y} * \frac{x(x - y)}{2(x^2 - y^2)} = \frac{x + y}{x - y} * \frac{x(x - y)}{2(x - y)(x + y)} = \frac{1}{1} * \frac{x}{2(x - y)} = \frac{x}{2(x - y)}$
$\frac{16 - m^2}{m^2 - 3m} : \frac{m^2 + 4m}{m^2 - 9} = \frac{16 - m^2}{m^2 - 3m} * \frac{m^2 - 9}{m^2 + 4m} = \frac{(4 - m)(4 + m)}{m(m - 3)} * \frac{(m - 3)(m + 3)}{m(m + 4)} = \frac{4 - m}{m} * \frac{m + 3}{m} = \frac{(4 - m)(m + 3)}{m^2}$
Пожауйста, оцените решение