8) $\frac{9}{12}+\frac{16}{20}$ и $\frac{3}{6}-\frac{2}{8}+\frac{6}{12}$

Question

Вопрос: $Фото задания: 8) $\frac{9}{12}+\frac{16}{20}$ и $\frac{3}{6}-\frac{2}{8}+\frac{6}{12}$$ Фото задания

Accepted Answer

$\frac{9}{12}+\frac{16}{20}=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}$.

$\frac{3}{6}-\frac{2}{8}+\frac{6}{12}=\frac{1}{2}-\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$.

Ответ: $\frac{9}{12}+\frac{16}{20}>\frac{3}{6}-\frac{2}{8}+\frac{6}{12}$.

8) \(\frac{9}{12}+\frac{16}{20}\) и \(\frac{3}{6}-\frac{2}{8}+\frac{6}{12}\)