1) $ \frac{(3.4)^{2}.4^{16}}{11.2^{13}.4^{11}} = \frac{3^{2}.4^{2}.4^{16}}{11.2^{13}.(2^{2})^{11}} = \frac{3^{2}.4^{18}}{11.2^{35}} = \frac{3^{2}.2}{11} = \frac{18}{11}$
3) $C = (\frac{239}{2004} + \frac{118}{1981} + \frac{92}{1986}).(\frac{1}{3} + \frac{1}{4} - \frac{7}{12})$
$= (\frac{239}{2004} + \frac{118}{1981} + \frac{92}{1986}).0 = 0$
4)
$\frac{25}{13} = 1 + \frac{12}{13} = 1 + \frac{120}{130}$
$\frac{257}{137} = 1 + \frac{120}{137}$
Do $1 + \frac{120}{130}$ > $1 + \frac{120}{137}$ nên $\frac{25}{13}$ > $\frac{257}{137}$