Cho \[M = \frac{{17}}{5}.\frac{{ - 31}}{{125}}.\frac{1}{2}.\frac{{10}}{{17}}.{\left( {\frac{{ - 1}}{2}} \right)^3}\] và \[N = \left( {\frac{{17}}{{28}} + \frac{{28}}{{29}} - \frac{{19}}{{30}} - \frac{{20}}{{31}}} \right).\left( {\frac{{ - 5}}{{12}} + \frac{1}{4} + \frac{1}{6}} \right)\]. Khi đó, tổng M + N bằng

Trả lời:

\[M = \frac{{17}}{5}.\frac{{ - 31}}{{125}}.\frac{1}{2}.\frac{{10}}{{17}}.{\left( {\frac{{ - 1}}{2}} \right)^3}\]

\[M = \frac{{17.\left( { - 31} \right).1.10.{{\left( { - 1} \right)}^3}}}{{{{5.125.2.17.2}^3}}}\]

\[M = \frac{{ - 31.\left( { - 1} \right)}}{{{{125.2}^3}}}\]

\[M = \frac{{31}}{{1000}}\]

\[N = \left( {\frac{{17}}{{28}} + \frac{{28}}{{29}} - \frac{{19}}{{30}} - \frac{{20}}{{31}}} \right).\left( {\frac{{ - 5}}{{12}} + \frac{1}{4} + \frac{1}{6}} \right)\]

\[N = \left( {\frac{{17}}{{28}} + \frac{{28}}{{29}} - \frac{{19}}{{30}} - \frac{{20}}{{31}}} \right).\left( {\frac{{ - 5}}{{12}} + \frac{3}{{12}} + \frac{2}{{12}}} \right)\]

\[N = \left( {\frac{{17}}{{28}} + \frac{{28}}{{29}} - \frac{{19}}{{30}} - \frac{{20}}{{31}}} \right).0\]

N = 0

Vậy \[M + N = \frac{{31}}{{1000}} + 0 = \frac{{31}}{{1000}}\]

Đáp án cần chọn là: B