\[\boxed{\mathbf{460\ (460).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
Пояснение.
Решение.
\[1)\ \left( \frac{5^{\backslash 10}}{6} + \frac{7^{\backslash 3}}{20} \right)\ :2 =\]
\[= \frac{50 + 21}{60}\ :2 = \frac{71}{60 \cdot 2} = \frac{71}{120}\]
\[3)\ \left( 2\frac{3}{5} + 3\frac{3}{10} + 2\frac{1}{2} \right)\ :3 =\]
\[= \left( \frac{13^{\backslash 2}}{5} + \frac{33}{10} + \frac{5^{\backslash 5}}{2} \right)\ :3 =\]
\[= \frac{26 + 33 + 25}{10} \cdot \frac{1}{3} = \frac{84}{10 \cdot 3} =\]
\[= \frac{14}{5} = 2\frac{4}{5}\]
\[4)\ \left( 7\frac{5}{24} + 6\frac{7}{24} + 8\frac{1^{\backslash 4}}{6} \right)\ :3 =\]
\[= 21\frac{5 + 7 + 4}{24}\ :3 = 21\frac{16}{24}\ :3 =\]
\[= 21\frac{2}{3}\ :3 = \frac{65}{3} \cdot \frac{1}{3} = \frac{65}{3 \cdot 3} =\]
\[= \frac{65}{9} = 7\frac{2}{9}\]
\[\boxed{\mathbf{460}.}\]
\[V_{пар}. = 1,2 \cdot 0,8 \cdot 1,5 =\]
\[= 1,2 \cdot 1,2 = 1,44\ дм^{3}.\]
\[V_{куба} = 1,1 \cdot 1,1 \cdot 1,1 =\]
\[= 1,21 \cdot 1,1 = 1,331\ дм^{3}.\]
\[Ответ:объем\]
\[\ параллелепипеда\ больше.\]