a) Do 2>1 và 24 > 16 => $2^{24} > 2^{16}$
b) Ta có: $(-\frac{1}{5})^{300} = [(-\frac{1}{5})^{3}]^{100} = (-\frac{1}{125})^{100} = (\frac{1}{125})^{100}$
$(-\frac{1}{3})^{500} = [(-\frac{1}{3})^{5}]^{100} = (-\frac{1}{243})^{100}= (\frac{1}{243})^{100}$
Do $\frac{1}{125} > \frac{1}{243} > 0$ nên $(\frac{1}{125})^{100} > (\frac{1}{243})^{100}. $
Vậy $(-\frac{1}{5})^{300} > (-\frac{1}{3})^{500}$
c) Do $\frac{32}{17} > 1 > 0 $nên $(\frac{32}{17})^{15} > 1.$ Mặt khác, $0 < \frac{17}{32} < 1$ nên$ (\frac{17}{32})^{30} < 1.$
Vậy $(\frac{32}{17})^{15} > (\frac{17}{32})^{30}.$