Докажите тождество:
$\frac{a^2 + a}{2a - 12} * \frac{6a + 6}{2a + 12} : \frac{9a^3 + 18a^2 + 9a}{a^2 - 36} = \frac{1}{6}$
$\frac{a^2 + a}{2a - 12} * \frac{6a + 6}{2a + 12} : \frac{9a^3 + 18a^2 + 9a}{a^2 - 36} = \frac{1}{6}$
$\frac{a(a + 1)}{2(a - 6)} * \frac{6(a + 1)}{2(a + 6)} : \frac{9a(a^2 + 2a + 1)}{(a - 6)(a + 6)} = \frac{1}{6}$
$\frac{a(a + 1)}{2(a - 6)} * \frac{3(a + 1)}{a + 6} : \frac{9a(a + 1)^2}{(a - 6)(a + 6)} = \frac{1}{6}$
$\frac{a(a + 1)}{2(a - 6)} * \frac{3(a + 1)}{a + 6} * \frac{(a - 6)(a + 6)}{9a(a + 1)^2} = \frac{1}{6}$
$\frac{1}{2} * \frac{1}{1} * \frac{1}{3} = \frac{1}{6}$
$\frac{1}{6} = \frac{1}{6}$
Пожауйста, оцените решение