Q 796. Представьте в виде суммы двух дробей с числителем 1 каждую из следующих дробей:$$\frac{5}{6}$$, $$\frac{8}{15}$$, $$\frac{9}{14}$$, $$\frac{4}{9}$$, $$\frac{13}{40}$$.

1. $$\frac{5}{6} = \frac{1}{2} + \frac{1}{3}$$ $$\frac{1}{2} + \frac{1}{3} = \frac{1 \cdot 3}{2 \cdot 3} + \frac{1 \cdot 2}{3 \cdot 2} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$$
2. $$\frac{8}{15} = \frac{1}{3} + \frac{1}{5}$$ $$\frac{1}{3} + \frac{1}{5} = \frac{1 \cdot 5}{3 \cdot 5} + \frac{1 \cdot 3}{5 \cdot 3} = \frac{5}{15} + \frac{3}{15} = \frac{8}{15}$$
3. $$\frac{9}{14} = \frac{1}{2} + \frac{1}{14}$$ $$\frac{1}{2} + \frac{1}{14} = \frac{1 \cdot 7}{2 \cdot 7} + \frac{1}{14} = \frac{7}{14} + \frac{2}{14} = \frac{9}{14}$$
4. $$\frac{4}{9} = \frac{1}{3} + \frac{1}{9}$$ $$\frac{1}{3} + \frac{1}{9} = \frac{1 \cdot 3}{3 \cdot 3} + \frac{1}{9} = \frac{3}{9} + \frac{1}{9} = \frac{4}{9}$$
5. $$\frac{13}{40} = \frac{1}{4} + \frac{3}{40}$$ $$\frac{1}{4} + \frac{3}{40} = \frac{1 \cdot 10}{4 \cdot 10} + \frac{3}{40} = \frac{10}{40} + \frac{3}{40} = \frac{13}{40}$$

Ответ: $$\frac{5}{6} = \frac{1}{2} + \frac{1}{3}$$, $$\frac{8}{15} = \frac{1}{3} + \frac{1}{5}$$, $$\frac{9}{14} = \frac{1}{2} + \frac{1}{14}$$, $$\frac{4}{9} = \frac{1}{3} + \frac{1}{9}$$, $$\frac{13}{40} = \frac{1}{4} + \frac{3}{40}$$