Tính: \(\,B = \frac{{\left( {\frac{{{8^2}}}{{128}} + \frac{3}{4}} \right):1\frac{3}{{16}}}}{{1\frac{{11}}{{19}}}}\)

* Hướng dẫn giải

\(\begin{array}{l}\,B = \frac{{\left( {\frac{{{8^2}}}{{128}} + \frac{3}{4}} \right):1\frac{3}{{16}}}}{{1\frac{{11}}{{19}}}} = \frac{{\left( {\frac{1}{2} + \frac{3}{4}} \right):\frac{{19}}{{16}}}}{{\frac{{30}}{{19}}}} = \frac{{\left( {\frac{2}{4} + \frac{3}{4}} \right).\frac{{16}}{{19}}}}{{\frac{{30}}{{19}}}}\\\,\,\,\,\,\,\,\,\, = \frac{{\frac{5}{4}.\frac{{16}}{{19}}}}{{\frac{{30}}{{19}}}} = \frac{{\frac{{20}}{{19}}}}{{\frac{{30}}{{19}}}} = \frac{{20}}{{19}}.\frac{{19}}{{30}} = \frac{2}{3}\end{array}\)

Chọn A