Представьте в виде дроби:
а) $\frac{mx^2 - my^2}{2m + 8} * \frac{3m + 12}{my + mx}$;
б) $\frac{ax + ay}{x^2 - 2xy + y^2} * \frac{x^2 - xy}{7x + 7y}$;
в) $\frac{x^3 - y^3}{x + y} * \frac{x^2 - y^2}{x^2 + xy + y^2}$;
г) $\frac{a^2 - 1}{a^3 + 1} * \frac{a^2 - a + 1}{a^2 + 2a + 1}$.
$\frac{mx^2 - my^2}{2m + 8} * \frac{3m + 12}{my + mx} = \frac{m(x - y)(x + y)}{2(m + 4)} * \frac{3(m + 4)}{m(y + x)} = \frac{x - y}{2} * \frac{3}{1} = \frac{3(x - y)}{2}$
$\frac{ax + ay}{x^2 - 2xy + y^2} * \frac{x^2 - xy}{7x + 7y} = \frac{a(x + y)}{(x - y)^2} * \frac{x(x - y)}{7(x + y)} = \frac{a}{x - y} * \frac{x}{7} = \frac{ax}{7(x - y)}$
$\frac{x^3 - y^3}{x + y} * \frac{x^2 - y^2}{x^2 + xy + y^2} = \frac{(x - y)(x^2 + xy + y^2)}{x + y} * \frac{(x - y)(x + y)}{x^2 + xy + y^2} = \frac{x - y}{1} * \frac{x - y}{1} = (x - y)^2$
$\frac{a^2 - 1}{a^3 + 1} * \frac{a^2 - a + 1}{a^2 + 2a + 1} = \frac{(a - 1)(a + 1)}{(a + 1)(a^2 - a + 1)} * \frac{a^2 - a + 1}{(a + 1)^2} = \frac{a - 1}{a + 1} * \frac{1}{a + 1} = \frac{a - 1}{(a + 1)^2}$
Пожауйста, оцените решение
