Решение:
- а) $$9\frac{5}{8} + 2\frac{1}{3}$$
\( 9\frac{5}{8} = \frac{9 \cdot 8 + 5}{8} = \frac{77}{8} \)
\( 2\frac{1}{3} = \frac{2 \cdot 3 + 1}{3} = \frac{7}{3} \)
\( \frac{77}{8} + \frac{7}{3} = \frac{77 \cdot 3 + 7 \cdot 8}{8 \cdot 3} = \frac{231 + 56}{24} = \frac{287}{24} = 11\frac{23}{24} \) - б) $$4\frac{3}{4} - 3\frac{5}{6}$$
\( 4\frac{3}{4} = \frac{4 \cdot 4 + 3}{4} = \frac{19}{4} \)
\( 3\frac{5}{6} = \frac{3 \cdot 6 + 5}{6} = \frac{23}{6} \)
\( \frac{19}{4} - \frac{23}{6} = \frac{19 \cdot 3 - 23 \cdot 2}{12} = \frac{57 - 46}{12} = \frac{11}{12} \) - в) $$6 - 3\frac{2}{9}$$
\( 6 = \frac{6}{1} = \frac{54}{9} \)
\( 3\frac{2}{9} = \frac{3 \cdot 9 + 2}{9} = \frac{29}{9} \)
\( \frac{54}{9} - \frac{29}{9} = \frac{25}{9} = 2\frac{7}{9} \)
Ответ: а) $$11\frac{23}{24}$$; б) $$\frac{11}{12}$$; в) $$2\frac{7}{9}$$.