$\text{a)}\left(\frac{14}{5}-\frac{21}{20}\right)\cdot \frac{4}{7}=\left(\frac{56}{20}-\frac{21}{20}\right)\cdot \frac{4}{7}=\frac{35}{20}\cdot \frac{4}{7}=\frac{7}{4}\cdot \frac{4}{7}=1$  

---

$\text{b)}\left(7\frac{1}{4}-2\frac{1}{6}\right):2\frac{13}{24}=\left(7\frac{3}{12}-2\frac{2}{12}\right):\frac{61}{24}=5\frac{1}{12}:\frac{61}{24}=\frac{61}{12}:\frac{61}{24}=\frac{{\sout{61}}^1}{{\sout{12}}^1}\cdot \frac{{\sout{24}}^2}{{\sout{61}}^1}=2$  

---

$\text{c)}\left(2\frac{4}{15}-1\frac{1}{10}\right)\cdot \left(\frac{2}{15}+\frac{2}{21}\right)=\left(2\frac{8}{30}-1\frac{3}{30}\right)\cdot \left(\frac{14}{105}+\frac{10}{105}\right)=1\frac{5}{30}\cdot \frac{24}{105}=$  

$=1\frac{1}{6}\cdot \frac{8}{35}=\frac{{\sout{7}}^1}{{\sout{6}}^3}\cdot \frac{{\sout{8}}^4}{{\sout{35}}^5}=\frac{4}{15}$  

---

$\text{d)}\frac{\frac{3}{4}:1\frac{1}{2}+4}{\left(6\frac{1}{2}-5\frac{3}{5}\right):\frac{9}{5}}-2=\frac{\frac{3}{4}:\frac{3}{2}+4}{\left(6\frac{5}{10}-5\frac{6}{10}\right):\frac{9}{5}}-2=\frac{\frac{{\sout{3}}^1}{{\sout{4}}^2}\cdot \frac{{\sout{2}}^1}{{\sout{3}}^1}+4}{\left(5\frac{15}{10}-5\frac{6}{10}\right)\cdot \frac{5}{9}}-2=$  

$=\frac{4\frac{1}{2}}{\frac{{\sout{9}}^1}{{\sout{10}}^2}\cdot \frac{{\sout{5}}^1}{{\sout{9}}^1}}-2=\frac{4\frac{1}{2}}{\frac{1}{2}}-2=4\frac{1}{2}:\frac{1}{2}-2=4\frac{1}{2}\cdot 2-2=9-2=7$  

---

$\text{e)}1,8\cdot \frac{3,3-4\frac{1}{2}:3\frac{3}{4}}{5,6:2\frac{1}{3}+2,5}=1,8\cdot \frac{3,3-\frac{9}{2}:\frac{15}{4}}{\frac{56}{10}:\frac{7}{3}+2\frac{5}{10}}=1,8\cdot \frac{3,3-\frac{{\sout{9}}^3}{{\sout{2}}^1}\cdot \frac{{\sout{4}}^2}{{\sout{15}}^5}}{\frac{{\sout{56}}^8}{10}\cdot \frac{3}{{\sout{7}}^1}+2\frac{5}{10}}=$  

$=\frac{18}{10}\cdot \frac{3\frac{3}{10}-\frac{6}{5}}{\frac{24}{10}+2\frac{5}{10}}=\frac{9}{5}\cdot \frac{3\frac{3}{10}-1\frac{1}{5}}{\frac{24}{10}+\frac{25}{10}}=\frac{9}{5}\cdot \frac{3\frac{3}{10}-1\frac{2}{10}}{\frac{49}{10}}=\frac{9}{5}\cdot \frac{2\frac{1}{10}}{\frac{49}{10}}=\frac{9}{5}\cdot \frac{\frac{21}{10}}{\frac{49}{10}}=$  

$=\frac{9}{5}\cdot \frac{21}{10}:\frac{49}{10}=\frac{9}{5}\cdot \frac{{\sout{21}}^3}{{\sout{10}}^1}\cdot \frac{{\sout{10}}^1}{{\sout{49}}^7}=\frac{9}{5}\cdot \frac{3}{7}=\frac{27}{35}$  

---

$\text{f)}2,125-\frac{\frac{4}{21}:\frac{2}{35}-3\frac{1}{9}\cdot \frac{3}{7}}{4\frac{4}{9}\cdot 0,4}=2\frac{125}{1000}-\frac{\frac{{\sout{4}}^2}{\sout{21}{}^3}\cdot \frac{{\sout{35}}^5}{{\sout{2}}^1}-\frac{{\sout{28}}^4}{{\sout{9}}^3}\cdot \frac{{\sout{3}}^1}{{\sout{7}}^1}}{\frac{{\sout{40}}^4}{9}\cdot \frac{4}{{\sout{10}}^1}}=2\frac{1}{8}-\frac{\frac{10}{3}-\frac{4}{3}}{\frac{16}{9}}=$  

$=2\frac{1}{8}-\frac{\frac{6}{3}}{\frac{16}{9}}=2\frac{1}{8}-\frac{2}{\frac{16}{9}}=2\frac{1}{8}-2:\frac{16}{9}=2\frac{1}{8}-{\sout{2}}^1\cdot \frac{9}{{\sout{16}}^8}=2\frac{1}{8}-\frac{9}{8}=$  

$=2\frac{1}{8}-1\frac{1}{8}=1$