\(\,B = \frac{{\left( {\frac{{{8^2}}}{{128}} + \frac{3}{4}} \right):1\frac{3}{{16}}}}{{1\frac{{11}}{{19}}}}\)
A.\(\frac{2}{3}\)
B.\(\frac{-2}{3}\)
C.\(\frac{1}{3}\)
D.\(\frac{-1}{3}\)

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