- \(\frac{4}{25}+\frac{15}{4}=\frac{16+375}{100}=\frac{391}{100}=3\frac{91}{100}\).
- \(\frac{9}{4}+\frac{8}{5}=\frac{45+32}{20}=\frac{77}{20}=3\frac{17}{20}\).
- \(\frac{3}{2}-\frac{9}{5}=\frac{15-18}{10}=-\frac{3}{10}\).
- \(\frac{3}{4}-\frac{9}{25}=\frac{75-36}{100}=\frac{39}{100}\).
- \(\frac{6}{5}\cdot\frac{3}{4}=\frac{18}{20}=\frac{9}{10}\).
- \(\frac{5}{3}\cdot\frac{9}{2}=\frac{45}{6}=7\frac{1}{2}\).
- \(\frac{12}{5}:\frac{15}{2}=\frac{12}{5}\cdot\frac{2}{15}=\frac{8}{25}\).
- \(\frac{21}{2}:\frac{3}{5}=\frac{21}{2}\cdot\frac{5}{3}=\frac{35}{2}=17\frac{1}{2}\).
- \(\frac{1}{\frac{1}{36}+\frac{1}{45}}=\frac{1}{\frac{5+4}{180}}=\frac{180}{9}=20\).
- \(\frac{1}{\frac{1}{35}-\frac{1}{60}}=\frac{1}{\frac{12-7}{420}}=\frac{420}{5}=84\).
- \(\left(\frac{3}{4}-\frac{1}{6}\right)\cdot3=\left(\frac{9-2}{12}\right)\cdot3=\frac{7}{4}=1\frac{3}{4}\).
- \(\left(\frac{2}{5}+\frac{13}{15}\right)\cdot6=\left(\frac{6+13}{15}\right)\cdot6=\frac{19}{15}\cdot6=\frac{38}{5}=7\frac{3}{5}\).
- \(\left(\frac{10}{13}+\frac{15}{4}\right)\cdot\frac{26}{5}=\left(\frac{40+195}{52}\right)\cdot\frac{26}{5}=\frac{235}{52}\cdot\frac{26}{5}=\frac{47}{2}=23\frac{1}{2}\).
- \(\left(\frac{5}{22}-\frac{8}{11}\right)\cdot\frac{11}{5}=\left(\frac{5-16}{22}\right)\cdot\frac{11}{5}=-\frac{1}{2}\).
- \(1\frac{8}{17}=\frac{25}{17}\), а \(12/17+2\frac{7}{11}=12/17+29/11=625/187\). Поэтому \(\frac{25}{17}:\frac{625}{187}=\frac{25}{17}\cdot\frac{187}{625}=\frac{11}{25}\).
- \(1\frac{1}{12}=\frac{13}{12}\), \(1\frac{13}{18}=\frac{31}{18}\), \(2\frac{5}{9}=\frac{23}{9}=\frac{46}{18}\). Тогда \(\frac{31}{18}-\frac{46}{18}=-\frac{5}{6}\), и \(\frac{13}{12}:\left(-\frac{5}{6}\right)=\frac{13}{12}\cdot\left(-\frac{6}{5}\right)=-\frac{13}{10}\).
Ответ: 1) \(3\frac{91}{100}\); 2) \(3\frac{17}{20}\); 3) \(-\frac{3}{10}\); 4) \(\frac{39}{100}\); 5) \(\frac{9}{10}\); 6) \(7\frac{1}{2}\); 7) \(\frac{8}{25}\); 8) \(17\frac{1}{2}\); 9) \(20\); 10) \(84\); 11) \(1\frac{3}{4}\); 12) \(7\frac{3}{5}\); 13) \(23\frac{1}{2}\); 14) \(-\frac{1}{2}\); 15) \(\frac{11}{25}\); 16) \(-\frac{13}{10}\).