Rozwiązanie:

1. a) $$4 \frac{3}{4} + 9 \frac{1}{4} = 4 + 9 + \frac{3}{4} + \frac{1}{4} = 13 + \frac{4}{4} = 13 + 1 = 14$$
2. b) $$2 \frac{2}{3} + 5 \frac{5}{6} = 2 + 5 + \frac{2}{3} + \frac{5}{6} = 7 + \frac{4}{6} + \frac{5}{6} = 7 + \frac{9}{6} = 7 + 1 \frac{3}{6} = 8 \frac{1}{2}$$
3. c) $$7 \frac{7}{12} - 3 \frac{1}{6} = 7 - 3 + \frac{7}{12} - \frac{1}{6} = 4 + \frac{7}{12} - \frac{2}{12} = 4 + \frac{5}{12} = 4 \frac{5}{12}$$
4. d) $$18 \frac{2}{5} - 9 \frac{9}{10} = 18 - 9 + \frac{2}{5} - \frac{9}{10} = 9 + \frac{4}{10} - \frac{9}{10} = 9 - \frac{5}{10} = 8 + \frac{10}{10} - \frac{5}{10} = 8 \frac{5}{10} = 8 \frac{1}{2}$$
5. e) $$4 \frac{4}{7} \cdot 2 \frac{5}{8} = \frac{32}{7} \cdot \frac{21}{8} = \frac{32 \cdot 21}{7 \cdot 8} = \frac{4 \cdot 3}{1 \cdot 1} = 12$$
6. f) $$\frac{3}{5} \cdot 4 = \frac{3 \cdot 4}{5} = \frac{12}{5} = 2 \frac{2}{5}$$
7. g) $$5 \frac{4}{9} : 2 \frac{13}{18} = \frac{49}{9} : \frac{49}{18} = \frac{49}{9} \cdot \frac{18}{49} = \frac{49 \cdot 18}{9 \cdot 49} = \frac{2}{1} = 2$$