Đáp án: 1
Giải thích các bước giải:
$\begin{array}{l}
A = \left[ {\dfrac{{1\dfrac{{11}}{{31}}.4\dfrac{3}{7} - \left( {15 - 6\dfrac{1}{3}.\dfrac{2}{{19}}} \right)}}{{4\dfrac{5}{6} + \dfrac{1}{6}\left( {12 - 5\dfrac{1}{3}} \right)}}.\left( { - 1\dfrac{{14}}{{93}}} \right)} \right].\dfrac{{31}}{{50}}\\
= \dfrac{{\dfrac{{42}}{{31}}.\dfrac{{31}}{7} - \left( {15 - \dfrac{{19}}{3}.\dfrac{2}{{19}}} \right)}}{{\dfrac{{29}}{6} + \dfrac{1}{6}.\left( {12 - \dfrac{{16}}{3}} \right)}}.\dfrac{{ - 107}}{{93}}.\dfrac{{31}}{{50}}\\
= \dfrac{{6 - 15 + \dfrac{2}{3}}}{{\dfrac{{29}}{6} + \dfrac{1}{6}.\dfrac{{20}}{3}}}.\dfrac{{ - 107}}{3}.\dfrac{1}{{50}}\\
= \dfrac{{\dfrac{{ - 25}}{3}}}{{\dfrac{{107}}{{18}}}}.\dfrac{{ - 107}}{3}.\dfrac{1}{{50}}\\
= \dfrac{{ - 25}}{3}.\dfrac{{18}}{{107}}.\dfrac{{ - 107}}{3}.\dfrac{1}{{50}}\\
= \dfrac{2}{2}\\
= 1
\end{array}$
