Câu 28-29. (1,0 điểm) Cho tia \(Ax.\) Trên tia \(Ax\) lấy điểm hai \(B\) và \[M\] sao cho  \[AB = 10\,{\rm{cm}}\] và \[AM = 2\,{\rm{cm}}{\rm{.}}\]

a) Tính độ dài đoạn thẳng \[MB.\]

1) a) \(1,25:\frac{{15}}{{20}} + \left( {25\% - \frac{5}{6}} \right):4\frac{2}{3}\)

\( = \frac{{125}}{{100}}:\frac{3}{4} + \left( {\frac{{25}}{{100}} - \frac{5}{6}} \right):\frac{{14}}{3}\)

\( = \frac{6}{5} \cdot \frac{4}{3} + \left( {\frac{1}{4} - \frac{5}{6}} \right) \cdot \frac{3}{{14}}\)

\( = \frac{8}{5} + \left( {\frac{3}{{12}} - \frac{{10}}{{12}}} \right) \cdot \frac{3}{{14}}\)

\( = \frac{8}{5} + \frac{{ - 7}}{{12}} \cdot \frac{3}{{14}}\)

\( = \frac{8}{5} + \frac{{ - 1}}{8} = \frac{{64}}{{40}} - \frac{5}{{40}} = \frac{{59}}{{40}}\).

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

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

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

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

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

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