a)

 \(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}}:\frac{3}{{ - 4}}} \right).\frac{4}{5} = \left( {\frac{2}{5}.\frac{{ - 4}}{3}} \right).\frac{4}{5}\\ = \frac{{ - 8}}{{15}}.\frac{4}{5} = \frac{{ - 32}}{{75}}\end{array}\)

b)

\(\begin{array}{l}\frac{{ - 3}}{{ - 4}}:\left( {\frac{7}{{ - 5}}.\frac{{ - 3}}{2}} \right) = \frac{3}{4}:\frac{{ - 21}}{{ - 10}}\\ = \frac{3}{4}.\frac{{10}}{{21}} = \frac{{30}}{{84}} = \frac{5}{14}\end{array}\)

c)

 \(\begin{array}{l}\frac{{ - 1}}{9}.\frac{{ - 3}}{5} + \frac{5}{{ - 6}}.\frac{{ - 3}}{5} + \frac{5}{2}.\frac{{ - 3}}{5}.\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 1}}{9} + \frac{5}{{ - 6}} + \frac{5}{2}} \right)\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 15}}{{18}} + \frac{{45}}{{18}}} \right)\\ = \frac{{ - 3}}{5}.\frac{{28}}{{18}}\\ = \frac{{ - 3}}{5}.\frac{{14}}{9}\\ = \frac{{ - 14}}{{15}}\end{array}\)

\(\frac{6:\frac{3}{5}-1\frac{1}{6}\cdot\frac{6}{7}}{4\frac{1}{5}\cdot\frac{10}{11}+5\frac{2}{11}}=1\)

mình cũng bằng 1

Phải là \(B=\frac{0,5-\frac{3}{17}+\frac{3}{37}}{\frac{5}{6}-\frac{5}{7}+\frac{5}{37}}+\frac{0,5-\frac{1}{3}+\frac{1}{4}-0,2}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-3,5}\)  chứ nhỉ?

Nếu đúng thì phân tích như sau

\(\Leftrightarrow B=\frac{\frac{3}{6}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{6}-\frac{5}{7}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)

\(B=\frac{3\left(\frac{1}{6}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{6}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)

\(B=\frac{3}{5}+\frac{1}{7}=\frac{16}{35}\)

\(\frac{6:\frac{3}{5}-1\frac{1}{6}\times\frac{6}{7}}{4\frac{1}{5}\times\frac{10}{11}+5\frac{2}{11}}\)

\(=\frac{\frac{6}{1}:\frac{3}{5}-\frac{7}{6}\times\frac{6}{7}}{\frac{21}{5}\times\frac{10}{11}\times\frac{57}{11}}\)

\(=\frac{\frac{6}{1}\times\frac{5}{3}-1}{\frac{210}{55}+\frac{57}{11}}\)

\(=\frac{\frac{30}{3}-1}{\frac{42}{11}+\frac{57}{11}}\)

\(=\frac{10-1}{\frac{99}{11}}\)

\(=\frac{9}{9}\)

\(=1\)

\(6:\frac{3}{5}-1\frac{1}{6}\)X \(\frac{6}{7}\)                                                     \(4\frac{1}{5}\)X \(\frac{10}{11}+5\frac{2}{11}\)

\(=\frac{33}{5}-\frac{7}{6}\)X \(\frac{6}{7}\)                                             \(=\)      \(\frac{21}{5}\)X  \(\frac{10}{11}+\frac{57}{11}\) 

\(=\frac{33}{5}-1\)                                                              \(=\frac{42}{11}+\frac{57}{11}\)

\(=\frac{28}{5}\)                                                                       \(=\frac{99}{11}=9\)

\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)

\(A=\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\left[\frac{6-\frac{43}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\frac{50}{31}\cdot\frac{31}{50}=1\)