PHẦN II. TỰ LUẬN

Thực hiện phép tính (tính hợp lí nếu có thể):

    a) \(1,25.6.\left( { - 8} \right)\);                  b) \(\frac{1}{8} - \frac{9}{8}.\frac{4}{3}\);

    c) \(1\frac{{13}}{{15}}.0,75 - \left( {\frac{{11}}{{20}} + 25\% } \right):\frac{5}{7}\);                                      d) \(\frac{{ - 5}}{{17}}.\frac{{31}}{{33}} - \frac{5}{{17}}.\frac{2}{{33}} + 1\frac{5}{{17}}\).

Ta có:

\(A = \frac{2}{{5.7}} + \frac{5}{{7.12}} + \frac{7}{{12.19}} + \frac{9}{{19.28}} + \frac{{11}}{{28.39}} + \frac{1}{{39.40}}\)

\( = \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{12}} + \frac{1}{{12}} - \frac{1}{{19}} + \frac{1}{{19}} - \frac{1}{{28}} + \frac{1}{{28}} - \frac{1}{{39}} + \frac{1}{{39}} - \frac{1}{{40}}\)

\( = \frac{1}{5} - \frac{1}{{40}}\)

\( = \frac{8}{{40}} - \frac{1}{{40}} = \frac{7}{{40}}\)

\[B = \frac{1}{{20}} + \frac{1}{{44}} + \frac{1}{{77}} + \frac{1}{{119}} + \frac{1}{{170}}\]

\[ = \frac{2}{{40}} + \frac{2}{{88}} + \frac{2}{{154}} + \frac{2}{{238}} + \frac{2}{{340}}\]

\[ = 2.\left( {\frac{1}{{5.8}} + \frac{1}{{8.11}} + \frac{1}{{11.14}} + \frac{1}{{14.17}} + \frac{1}{{17.20}}} \right)\]

\[ = 2.\frac{1}{3}.\left( {\frac{3}{{5.8}} + \frac{3}{{8.11}} + \frac{3}{{11.14}} + \frac{3}{{14.17}} + \frac{3}{{17.20}}} \right)\]

\[ = \frac{2}{3}.\left( {\frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{{14}} + \frac{1}{{14}} - \frac{1}{{17}} + \frac{1}{{17}} - \frac{1}{{20}}} \right)\]

\[ = \frac{2}{3}.\left( {\frac{1}{5} - \frac{1}{{20}}} \right)\]

\[ = \frac{2}{3}.\left( {\frac{4}{{20}} - \frac{1}{{20}}} \right)\]

\[ = \frac{2}{3}.\frac{3}{{20}}\]

\[ = \frac{1}{{10}}\].

Ta có: \(\frac{7}{{40}} > \frac{4}{{40}} = \frac{1}{{10}}\)

Do đó \(A > B\).