Решебник по математике 6 класс Мерзляк ФГОС Задание 241

\[\boxed{\mathbf{241\ (241)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

Пояснение.

Решение.

\[1)\ \frac{5}{7} = \frac{45}{63};\ \ \ \ \ \frac{7}{9} = \frac{49}{63} \Longrightarrow \frac{5}{7} < \frac{7}{9}\]

\[2)\ \frac{11}{20} = \frac{33}{60};\ \ \ \ \frac{17}{30} = \frac{34}{60} \Longrightarrow\]

\[\Longrightarrow \frac{11}{20} < \frac{17}{30}\]

\[3)\ \frac{2}{9} = \frac{4}{18};\ \ \ \ \frac{1}{6} = \frac{3}{18} \Longrightarrow \frac{2}{9} > \frac{1}{6}\]

\[4)\ \frac{5}{6} = \frac{10}{12};\ \ \ \frac{3}{4} = \frac{9}{12} \Longrightarrow \frac{5}{6} > \frac{3}{4}\]

\[5)\ \frac{8}{38};\ \ \ \frac{4}{19} = \frac{8}{38} \Longrightarrow \frac{8}{38} = \frac{4}{19}\]

\[6)\ \frac{7}{9} = \frac{77}{99};\ \ \ \ \frac{8}{11} = \frac{72}{99} \Longrightarrow \frac{7}{9} > \frac{8}{11}\]

\[7)\ \frac{8}{25} = \frac{32}{100};\ \ \ \ \frac{7}{20} = \frac{35}{100} \Longrightarrow\]

\[\Longrightarrow \frac{32}{100} < \frac{35}{100}\]

\[8)\ \frac{5}{12} = \frac{15}{36};\ \ \ \ \ \ \frac{4}{9} = \frac{16}{36} \Longrightarrow\]

\[\Longrightarrow \frac{5}{12} < \frac{4}{9}\]

\[\boxed{\mathbf{241}\mathbf{.}}\]

\[7 \cdot 7 = 49\ м^{2} = 490\ 000\ см^{2} -\]

\[площадь\ квадрата.\]

\[Ответ:490\ 000\ см^{2}.\]