a)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A =  - 1\end{array}\)

b)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A =  - 1 + 0 + 0 =  - 1\end{array}\)

a)

\(\begin{array}{l}\frac{2}{3} + \frac{{ - 2}}{5} + \frac{{ - 5}}{6} - \frac{{13}}{{10}}\\ = \frac{2}{3} + \frac{{ - 5}}{6} + \frac{{ - 2}}{5} - \frac{{13}}{{10}}\\ = \left( {\frac{2}{3} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 2}}{5} - \frac{{13}}{{10}}} \right)\\ = \left( {\frac{4}{6} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 4}}{{10}} - \frac{{13}}{{10}}} \right)\\ = \frac{{ - 1}}{6} + \frac{{ - 17}}{{10}}\\ = \frac{{ - 5}}{{30}} + \frac{{ - 51}}{{30}}\\ = \frac{{ - 56}}{{30}}\\ = \frac{{ - 28}}{{15}}\end{array}\)

b)

\(\begin{array}{l}\frac{{ - 3}}{7}.\frac{{ - 1}}{9} + \frac{7}{{ - 18}}.\frac{{ - 3}}{7} + \frac{5}{6}.\frac{{ - 3}}{7}\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 1}}{9} + \frac{7}{{ - 18}} + \frac{5}{6}} \right)\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 7}}{{18}} + \frac{{15}}{{18}}} \right)\\ = \frac{{ - 3}}{7}.\frac{{ 6}}{{18}}\\ = \frac{-1}{7}\end{array}\).

\(\Rightarrow A=\frac{\frac{1}{2}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{4}+\frac{1}{2}+\frac{3}{10}}\)

\(\Rightarrow A=\frac{\frac{1}{2}-\frac{1}{5}+\frac{1}{7}}{3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{7}\right)}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{2}.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\right)}\)

\(\Rightarrow A=\frac{1}{3}+\frac{2}{3}\)

\(\Rightarrow A=1\)

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

mình cũng bằng 1

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

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

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

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

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

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

\(=1\)

\(6:\frac{3}{5}-1\frac{1}{6}\)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  \(\frac{10}{11}+\frac{57}{11}\) 

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

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

a) Cách 1:

\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = 8 + \frac{7}{3} - \frac{3}{5} - 5 - \frac{2}{5} - \frac{{10}}{3} + 2\\ = (8 - 5 + 2) + (\frac{7}{3} - \frac{{10}}{3}) - (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ - 3}}{3} - \frac{5}{5}\\ = 5 + ( - 1) - 1\\ = 3\end{array}\)

Cách 2:

\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} - \frac{9}{{15}}) - (\frac{{25}}{5} + \frac{2}{5}) - (\frac{{10}}{3} - \frac{6}{3})\\ = \frac{{146}}{{15}} - \frac{{27}}{5} - \frac{4}{3}\\ = \frac{{146}}{{15}} - \frac{{81}}{{15}} - \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)

b)

\(\begin{array}{l}(7 - \frac{1}{2} - \frac{3}{4}):(5 - \frac{1}{4} - \frac{5}{8})\\ = (\frac{{28}}{4} - \frac{2}{4} - \frac{3}{4}):(\frac{{40}}{8} - \frac{2}{8} - \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)