\[\boxed{\text{169\ (169).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\frac{a^{2} + a}{2a - 12} \cdot \frac{6a + 6}{2a + 12}\ :\]
\[:\frac{9a^{3} + 18a^{2} + 9a}{a^{2} - 36} = \frac{1}{6}\]
\[Упростим\ левую\ часть\ \]
\[тождества:\]
\[\frac{a(a + 1) \cdot 6(a + 1) \cdot (a - 6)(a + 6)}{2(a - 6) \cdot 2(a + 6) \cdot 9a(a + 1)^{2}} =\]
\[= \frac{1}{6}\]
\[\frac{3}{18} = \frac{1}{6}\]
\[\frac{1}{6} = \frac{1}{6}.\]
\[Тождество\ доказано.\]