Решебник по алгебре 7 класс Мерзляк ФГОС Задание 279

\[\boxed{\text{279\ (279).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ 81m^{4}n^{16} =\]

\[= - \frac{1}{3}mn^{14} \cdot ( - 243m^{3}n^{2})\]

\[2)\ 81m^{4}n^{16} =\]

\[= \left( 9m^{2}n^{8} \right)^{2}\ или\ \ \left( \ - 9m^{2}n^{8} \right)^{2}\]

\[3)\ 81m^{4}n^{16} =\]

\[= \left( 3mn^{4} \right)^{4}\ \ или\ \left( - 3mn^{4} \right)^{4}\]

\[\boxed{\text{279.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ 3\frac{1}{3} \cdot 1,3 = \frac{10}{3} \cdot \frac{13}{10} = \frac{13}{3}.\]

\[2)\ 7,2 \cdot \frac{2}{27} = \frac{72}{10} \cdot \frac{2}{27} = \frac{72 \cdot 2}{10 \cdot 27} =\]

\[= \frac{8 \cdot 1}{5 \cdot 3} = \frac{8}{15}.\]

\[3)\ 9,1\ :3,5 = \frac{91}{10}\ :\frac{35}{10} =\]

\[= \frac{91 \cdot 10}{10 \cdot 35} = \frac{13 \cdot 1}{1 \cdot 5} = \frac{13}{5}.\]

\[4,5)\ \frac{13^{\backslash 5}}{3} - \frac{8}{15} - \frac{13^{\backslash 3}}{5} =\]

\[= \frac{65 - 8 - 39}{15} = \frac{18}{15} = \frac{6}{5}.\]

\[6)\ \frac{6}{5}\ :\frac{2}{5} = \frac{6 \cdot 5}{5 \cdot 2} = \frac{3 \cdot 1}{1 \cdot 1} = 3.\ \]