$3^{-2}$ и $(-3)^{0}$
$3^{-2} = (\frac{1}{3})^2 = \frac{1}{9}$
$(-3)^{0} = 1$
$\frac{1}{9} < 1$
Ответ:
$3^{-2} < (-3)^{0}$

$3^{-1} + 2^{-1}$ и $5^{-1}$
$3^{-1} + 2^{-1} = \frac{1}{3} + \frac{1}{2} = \frac{2 + 3}{6} = \frac{5}{6} = \frac{25}{30}$
$5^{-1} = \frac{1}{5} = \frac{6}{30}$
$\frac{25}{30} > \frac{6}{30}$
Ответ:
$3^{-1} + 2^{-1} > 5^{-1}$

$(\frac{1}{4})^{-2} - (\frac{1}{5})^{-2}$ и $(\frac{1}{4} - \frac{1}{5})^{-2}$
$(\frac{1}{4})^{-2} - (\frac{1}{5})^{-2} = 4^2 - 5^2 = 16 - 25 = -9$
$(\frac{1}{4} - \frac{1}{5})^{-2} = (\frac{5 - 4}{20})^{-2} = (\frac{1}{20})^{-2} = 20^2 = 400$
−9 < 400
Ответ:
$(\frac{1}{4})^{-2} - (\frac{1}{5})^{-2} < (\frac{1}{4} - \frac{1}{5})^{-2}$