Chọn dấu "<", ">", "=" thích hợp vào chỗ trống

$a) \frac{5}{6} - (\frac{1}{6})^{2} = \frac{5}{6} - \frac{1}{36} = \frac{29}{36}$

$(\frac{5}{6} - \frac{1}{6})^{2} = (\frac{2}{3})^{2} = \frac{4}{9}$

 $\frac{29}{36} > \frac{4}{9}$

$=> \frac{5}{6} - (\frac{1}{6})^{2} > (\frac{5}{6} - \frac{1}{6})^{2};$

$b) 250 \times (\frac{1}{30})^{2} = 250 \times \frac{1}{900} = \frac{5}{18}$

$250 \times (\frac{1}{5})^{2} - \frac{1}{6} = 250 \times \frac{1}{25} - \frac{1}{6} = 10 - \frac{1}{6} = \frac{59}{6}$

$\frac{5}{18} < \frac{59}{6}$

$250 \times (\frac{5}{6} - \frac{1}{6})^{2} < 250 \times (\frac{1}{5})^{2} - \frac{1}{6};$

$c) 3\frac{1}{5} / 1.5 + 4\frac{2}{5} /1.5 = (3\frac{1}{5} + 4\frac{2}{5}) / 1.5;$

$d)(\frac{9}{25} - 2.18) / (3\frac{4}{5}) + 0.2) = \frac{-91}{50} / 4 = \frac{-91}{200}$

$\frac{9}{25} / 3\frac{4}{5} - 2.18 / 0.2 = \frac{9}{95} - \frac{109}{10} = -\frac{2053}{190}$

$\frac{-91}{200} > -\frac{2053}{190}$

$=> (\frac{9}{25} - 2.18) / (3\frac{4}{5}) + 0.2) > \frac{9}{25} / 3\frac{4}{5} - 2.18 / 0.2.$