Решебник по математике 6 класс Мерзляк ФГОС Задание 349

\[\boxed{\mathbf{349\ (349).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \left( \frac{1}{2} \right)^{4} = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \cdot \frac{1}{4} =\]

\[= \frac{1}{16}\]

\[2)\ \left( \frac{2}{5} \right)^{3} = \frac{2}{5} \cdot \frac{2}{5} \cdot \frac{2}{5} = \frac{4 \cdot 2}{25 \cdot 5} =\]

\[= \frac{8}{125}\]

\[3)\ \left( 1\frac{1}{3} \right)^{4} = \left( \frac{4}{3} \right)^{4} = \frac{4}{3} \cdot \frac{4}{3} \cdot \frac{4}{3} \cdot \frac{4}{3} =\]

\[= \frac{16}{9} \cdot \frac{16}{9} = \frac{256}{81} = 3\frac{13}{81}\]

\[4)\ \left( 2\frac{2}{3} \right)^{2} = \left( \frac{8}{3} \right)^{2} = \frac{8}{3} \cdot \frac{8}{3} = \frac{64}{9} =\]

\[= 7\frac{1}{9}\]

\[\boxed{\mathbf{349}\mathbf{.}}\]

\[1)\ 2\frac{5}{8} \cdot \frac{2}{3} + \frac{7}{8} = \frac{21}{8} \cdot \frac{2}{3} + \frac{7}{8} =\]

\[= \frac{7 \cdot 2}{8} + \frac{7}{8} = \frac{14}{8} + \frac{7}{8} = \frac{21}{8} = 2\frac{5}{8}\]

\[2)\ \left( 6 - 7\frac{7}{9} \cdot \frac{3}{5} \right) \cdot \frac{7}{12} =\]

\[= 6 \cdot \frac{7}{12} - \frac{70}{9} \cdot \frac{3}{5} \cdot \frac{7}{12} =\]

\[= \frac{7}{2} - \frac{14 \cdot 7}{3 \cdot 12} = \frac{7}{2} - \frac{49}{18} =\]

\[= 3\frac{1^{\backslash 9}}{2} - 2\frac{13}{18} = 3\frac{9}{18} - 2\frac{13}{18} =\]

\[= 2\frac{27}{18} - 2\frac{13}{18} = \frac{14}{18} = \frac{7}{9}\]