Преобразуйте в алгебраическую дробь:
а) $\frac{m^3 + n^3}{2m} * \frac{4mn}{m^2 - mn + n^2}$;
б) $\frac{2a}{a^3 - b^3} : \frac{6ab}{a^2 - b^2}$;
в) $\frac{m^3 - n^3}{m^3 + n^3} : \frac{(m - n)^2}{m^2 - n^2}$;
г) $\frac{x^2 + xy}{6x^2 - 6y^2} * \frac{3x^3 + 3y^3}{x^2 - xy}$;
д) $\frac{p^2 - 4q^2}{(p + 2q)^2} : \frac{p^3 - 8q^3}{4q^2 + 2pq + p^2}$;
е) $\frac{12a^2 + 6ab}{8a^3 - b^3} * \frac{4a^2 + 2ab + b^2}{3a^2 - 6ab}$.
$\frac{m^3 + n^3}{2m} * \frac{4mn}{m^2 - mn + n^2} = \frac{(m + n)(m^2 - mn + n^2)}{2m} * \frac{4mn}{m^2 - mn + n^2} = \frac{m + n}{1} * \frac{2n}{1} = 2n(m + n)$
$\frac{2a}{a^3 - b^3} : \frac{6ab}{a^2 - b^2} = \frac{2a}{a^3 - b^3} * \frac{a^2 - b^2}{6ab} = \frac{2a}{(a - b)(a^2 + ab + b^2)} * \frac{(a - b)(a + b)}{6ab} = \frac{1}{a^2 + ab + b^2} * \frac{a + b}{3b} = \frac{a + b}{3b(a^2 + ab + b^2)}$
$\frac{m^3 - n^3}{m^3 + n^3} : \frac{(m - n)^2}{m^2 - n^2} = \frac{m^3 - n^3}{m^3 + n^3} * \frac{m^2 - n^2}{(m - n)^2} = \frac{(m - n)(m^2 + mn + n^2)}{(m + n)(m^2 - mn + n^2)} * \frac{(m - n)(m + n)}{(m - n)^2} = \frac{(m - n)(m^2 + mn + n^2)}{(m + n)(m^2 - mn + n^2)} * \frac{m + n}{m - n} = \frac{m^2 + mn + n^2}{m^2 - mn + n^2}$
$\frac{x^2 + xy}{6x^2 - 6y^2} * \frac{3x^3 + 3y^3}{x^2 - xy} = \frac{x(x + y)}{6(x^2 - y^2)} * \frac{3(x^3 + y^3)}{x(x - y)} = \frac{x(x + y)}{6(x - y)(x + y)} * \frac{3(x + y)(x^2 - xy + y^2)}{x(x - y)} = \frac{x + y}{2(x - y)} * \frac{x^2 - xy + y^2}{x - y} = \frac{(x + y)(x^2 - xy + y^2)}{2(x - y)(x - y)} = \frac{x^3 + y^3}{2(x - y)^2}$
$\frac{p^2 - 4q^2}{(p + 2q)^2} : \frac{p^3 - 8q^3}{4q^2 + 2pq + p^2} = \frac{p^2 - 4q^2}{(p + 2q)^2} * \frac{4q^2 + 2pq + p^2}{p^3 - 8q^3} = \frac{(p - 2q)(p + 2q)}{p^2 + 4pq + 4q^2} * \frac{4q^2 + 2pq + p^2}{(p - 2q)(p^2 + 2pq + 4q^2)} = \frac{p + 2q}{1} * \frac{1}{p^2 + 2pq + 4q^2} = \frac{p + 2q}{(p + 2q)^2} = \frac{1}{p + 2q}$
$\frac{12a^2 + 6ab}{8a^3 - b^3} * \frac{4a^2 + 2ab + b^2}{3a^2 - 6ab} = \frac{6a(2a + b)}{(2a - b)(4a^2 + 2ab + b^2)} * \frac{4a^2 + 2ab + b^2}{3a(a - 2b)} = \frac{2(2a + b)}{2a - b} * \frac{1}{a - 2b} = \frac{2(2a + b)}{(2a - b)(2a - b)}$
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