Представьте в виде неправильных дробей смешанные числа: 3 2/5; 1 5/6; 3 8/11; 1 51/53; 7 3/7; 17 2/3; 5 2/5; 3 5/12; 3 3/4; 2 1/4; 2 8/9; 10 2/3.

Используем правило: \(a\frac{b}{c}=\frac{ac+b}{c}\).

1. \(3\frac{2}{5}=\frac{3\cdot5+2}{5}=\frac{17}{5}\);
2. \(1\frac{5}{6}=\frac{1\cdot6+5}{6}=\frac{11}{6}\);
3. \(3\frac{8}{11}=\frac{3\cdot11+8}{11}=\frac{41}{11}\);
4. \(1\frac{51}{53}=\frac{1
   a\cdot53+51}{53}=\frac{104}{53}\);
5. \(7\frac{3}{7}=\frac{7\cdot7+3}{7}=\frac{52}{7}\);
6. \(17\frac{2}{3}=\frac{17\cdot3+2}{3}=\frac{53}{3}\);
7. \(5\frac{2}{5}=\frac{5\cdot5+2}{5}=\frac{27}{5}\);
8. \(3\frac{5}{12}=\frac{3\cdot12+5}{12}=\frac{41}{12}\);
9. \(3\frac{3}{4}=\frac{3\cdot4+3}{4}=\frac{15}{4}\);
10. \(2\frac{1}{4}=\frac{2\cdot4+1}{4}=\frac{9}{4}\);
11. \(2\frac{8}{9}=\frac{2\cdot9+8}{9}=\frac{26}{9}\);
12. \(10\frac{2}{3}=\frac{10\cdot3+2}{3}=\frac{32}{3}\).

Ответ: \(\frac{17}{5};\frac{11}{6};\frac{41}{11};\frac{104}{53};\frac{52}{7};\frac{53}{3};\frac{27}{5};\frac{41}{12};\frac{15}{4};\frac{9}{4};\frac{26}{9};\frac{32}{3}\).