(0,5 điểm) Cho hình bình hành \(ABCD.\) Vẽ hình đối xứng qua tâm là đỉnh \(A\) của hình bình hành \(ABCD.\)

1) a) \[\left( {\frac{4}{5} + \frac{{ - 9}}{7}} \right):\frac{2}{3} + \left( {\frac{{ - 5}}{7} - \frac{{ - 6}}{5}} \right):\frac{2}{3}\]

\[ = \left( {\frac{4}{5} + \frac{{ - 9}}{7} + \frac{{ - 5}}{7} - \frac{{ - 6}}{5}} \right):\frac{2}{3}\]

\[ = \left[ {\left( {\frac{4}{5} - \frac{{ - 6}}{5}} \right) + \left( {\frac{{ - 9}}{7} + \frac{{ - 5}}{7}} \right)} \right]:\frac{2}{3}\]

\[ = \left[ {\frac{{10}}{5} + \frac{{ - 14}}{7}} \right]:\frac{2}{3}\]

\[ = \left[ {2 + \left( { - 2} \right)} \right]:\frac{2}{3}\]

\[ = 0:\frac{2}{3} = 0.\]

b) \(\left( { - 2,4 + \frac{1}{3}} \right):3,1 + 75\% :1\frac{1}{2}\)

\[ = \left( { - \frac{{24}}{{10}} + \frac{1}{3}} \right):\frac{{31}}{{10}} + \frac{{75}}{{100}}:\frac{3}{2}\]

\[ = \left( { - \frac{{12}}{5} + \frac{1}{3}} \right) \cdot \frac{{10}}{{31}} + \frac{3}{4} \cdot \frac{2}{3}\]

\[ = \left( { - \frac{{36}}{{15}} + \frac{5}{{15}}} \right) \cdot \frac{{10}}{{31}} + \frac{1}{2}\]

\[ = \frac{{ - 31}}{{15}} \cdot \frac{{10}}{{31}} + \frac{1}{2}\]\[ = \frac{{ - 2}}{3} + \frac{1}{2}\]

\[ = \frac{{ - 4}}{6} + \frac{3}{6} = - \frac{1}{6}.\]