a) $\mathrm{A}=\left[\genfrac{}{}{0ex}{}{2}{7}\left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{3}\right)\right]:\left[\genfrac{}{}{0ex}{}{2}{7}\left(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{2}{5}\right)\right]=\left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{3}\right):\left(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{2}{5}\right)=1\genfrac{}{}{0ex}{}{1}{4}$.
b) $\mathrm{B}=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{3}{4}\left(\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{2}{7}-\genfrac{}{}{0ex}{}{1}{3}+\genfrac{}{}{0ex}{}{2}{7}\right)}{\genfrac{}{}{0ex}{}{1}{5}\left(\genfrac{}{}{0ex}{}{2}{7}+\genfrac{}{}{0ex}{}{1}{3}\right)-\genfrac{}{}{0ex}{}{1}{3}\left(\genfrac{}{}{0ex}{}{2}{7}+\genfrac{}{}{0ex}{}{1}{3}\right)}=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{3}{4}\left(\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{1}{3}\right)}{\left(\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{1}{3}\right)\left(\genfrac{}{}{0ex}{}{2}{7}+\genfrac{}{}{0ex}{}{1}{3}\right)}=1\genfrac{}{}{0ex}{}{11}{52}$.

a) $\mathrm{A}=\genfrac{}{}{0ex}{}{3}{5}.\genfrac{}{}{0ex}{}{6}{7}+\genfrac{}{}{0ex}{}{3}{7}.\genfrac{}{}{0ex}{}{3}{5}-\genfrac{}{}{0ex}{}{2}{7}.\genfrac{}{}{0ex}{}{3}{5}$
$=\genfrac{}{}{0ex}{}{3}{5}\cdot \left(\genfrac{}{}{0ex}{}{6}{7}+\genfrac{}{}{0ex}{}{3}{7}-\genfrac{}{}{0ex}{}{2}{7}\right)=\genfrac{}{}{0ex}{}{3}{5}$
b) $\mathrm{B}=\left(-13\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{-2}{9}\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{2}{5}\cdot \genfrac{}{}{0ex}{}{11}{9}\right)\cdot \genfrac{}{}{0ex}{}{5}{2}$
$=\left(-13-\genfrac{}{}{0ex}{}{2}{9}+\genfrac{}{}{0ex}{}{11}{9}\right)\cdot \genfrac{}{}{0ex}{}{2}{5}\cdot \genfrac{}{}{0ex}{}{5}{2}=-13+\left(\genfrac{}{}{0ex}{}{11}{9}-\genfrac{}{}{0ex}{}{2}{9}\right)=-12.$
c) $\mathrm{C}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}+\left(\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=\left(\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{-1}{5}\right)+\left(\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{2}{7}\right)\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=0.$
d) $\mathrm{D}=\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{1}{15}-\genfrac{}{}{0ex}{}{10}{15}\right)+\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{2}{22}-\genfrac{}{}{0ex}{}{5}{22}\right)$
$=\genfrac{}{}{0ex}{}{4}{9}:\genfrac{}{}{0ex}{}{-3}{5}+\genfrac{}{}{0ex}{}{4}{9}:\genfrac{}{}{0ex}{}{-3}{22}=\genfrac{}{}{0ex}{}{4}{9}\cdot \genfrac{}{}{0ex}{}{-5}{3}+\genfrac{}{}{0ex}{}{4}{9}.\genfrac{}{}{0ex}{}{-22}{3}$
$=\genfrac{}{}{0ex}{}{4}{9}\cdot \left(\genfrac{}{}{0ex}{}{-5}{3}+\genfrac{}{}{0ex}{}{-22}{3}\right)=\genfrac{}{}{0ex}{}{4}{9}.\genfrac{}{}{0ex}{}{-27}{3}=-4.$

$\mathrm{a)\; P}=\frac{2}{3}+\frac{1}{4}+\frac{3}{5}-\frac{7}{45}+\frac{5}{9}+\frac{1}{12}+\frac{1}{35}$ 

    P $=\left(\frac{2}{3}+\frac{1}{4}+\frac{1}{12}\right)+\left(\frac{5}{9}-\frac{7}{45}\right)+\frac{3}{5}+\frac{1}{35}$

P =$1+\frac{4}{5}+\frac{3}{5}+\frac{1}{35}=2\frac{1}{35}$.

b)

$Q\; =\left(5-\frac{3}{4}+\frac{1}{5}\right)-\left(6+\frac{7}{4}-\frac{8}{5}\right)-\left(2-\frac{5}{4}+\frac{16}{5}\right)$

$Q=(5-6-2)+\left(-\frac{3}{4}-\frac{7}{4}+\frac{5}{4}\right)+(\frac{1}{5}-\frac{8}{5}-\frac{16}{5}$)

$Q\; =\; -\left(3+\frac{5}{4}+\frac{23}{5}\right)$ $=-(3+1\frac{1}{4}+4\frac{3}{5}$

$Q\; =-8\frac{17}{20}$.

a) $A=\left(\frac{1}{3}+\frac{2}{3}\right)-\left(\frac{8}{15}+\frac{7}{15}\right)+\left(\frac{-1}{7}+1\frac{1}{7}\right)=1-1+1=1$;
b) $B=\left(0.25-1\frac{1}{4}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)-\frac{1}{8}$
$=\left(\frac{1}{4}-1-\frac{1}{4}\right)+1-\frac{1}{8}=\frac{-1}{8}$.