Điền dấu >, <, = thích hợp vào chỗ chấm

a) \[\frac{3}{5} + \frac{9}{{10}}\,\,...\,\,\frac{5}{2} + \frac{3}{4}\]               b) \[\frac{{11}}{4} - \frac{1}{2}\,\,...\,\,\frac{5}{4} \times \frac{3}{8}\]

c) \[\frac{5}{9} \times \frac{2}{7}\,\,...\,\,\frac{7}{3}:\frac{3}{2}\]               d) \[\frac{9}{8}:\frac{3}{4}\,\,...\,\,\frac{1}{6} + \frac{2}{3}\]

e) \[\frac{{10}}{3} - \frac{1}{4}\,\,...\,\,\frac{5}{6} + \frac{1}{{12}}\]           f) \[\frac{5}{8} + \frac{3}{4}\,\,...\,\,\frac{1}{6} + \frac{{11}}{{24}}\]

a)

\[\frac{3}{5} + \frac{9}{{10}} = \frac{6}{{10}} + \frac{9}{{10}} = \frac{{15}}{{10}} = \frac{{30}}{{20}}\]

\[\frac{5}{2} + \frac{3}{4} = \frac{{10}}{4} + \frac{3}{4} = \frac{{13}}{4} = \frac{{65}}{{20}}\]

So sánh: \[\frac{{30}}{{20}} < \frac{{65}}{{20}}\]

Vậy: \[\frac{3}{5} + \frac{9}{{10}}\,\, < \,\,\,\frac{5}{2} + \frac{3}{4}\]

b)

\[\frac{{11}}{4} - \frac{1}{2} = \frac{{11}}{4} - \frac{2}{4} = \frac{9}{4} = \frac{{72}}{{32}}\]

\[\frac{5}{4} \times \frac{3}{8} = \frac{{15}}{{32}}\]

So sánh: \[\frac{{72}}{{32}} > \frac{{15}}{{32}}\]

Vậy: \[\frac{{11}}{4} - \frac{1}{2}\,\, > \,\,\frac{5}{4} \times \frac{3}{8}\]

c)

\[\frac{5}{9} \times \frac{2}{7} = \frac{{10}}{{63}}\]

\[\frac{7}{3}:\frac{3}{2} = \frac{7}{3} \times \frac{2}{3} = \frac{{14}}{9} = \frac{{98}}{{63}}\]

So sánh: \[\frac{{10}}{{63}} < \frac{{98}}{{63}}\]

Vậy: \[\frac{5}{9} \times \frac{2}{7}\,\, < \,\,\frac{7}{3}:\frac{3}{2}\]

d)

\[\frac{9}{8}:\frac{3}{4} = \frac{9}{8} \times \frac{4}{3} = \frac{9}{6}\]

\[\frac{1}{6} + \frac{2}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6}\]

So sánh: \[\frac{9}{6} > \frac{5}{6}\]

Vậy: \[\frac{9}{8}:\frac{3}{4}\,\, > \,\,\frac{1}{6} + \frac{2}{3}\]

e)

\[\frac{{10}}{3} - \frac{1}{4} = \frac{{40}}{{12}} - \frac{3}{{12}} = \frac{{37}}{{12}}\]

\[\frac{5}{6} + \frac{1}{{12}} = \frac{{10}}{{12}} + \frac{1}{{12}} = \frac{{11}}{{12}}\]

So sánh: \[\frac{{37}}{{12}} > \frac{{11}}{{12}}\]

Vậy: \[\frac{{10}}{3} - \frac{1}{4}\,\, > \,\,\frac{5}{6} + \frac{1}{{12}}\]          

f)

\[\frac{5}{8} + \frac{3}{4} = \frac{5}{8} + \frac{6}{8} = \frac{{11}}{8} = \frac{{33}}{{24}}\]

\[\frac{1}{6} + \frac{{11}}{{24}} = \frac{4}{{24}} + \frac{{11}}{{24}} = \frac{{15}}{{24}}\]

So sánh: \[\frac{{33}}{{24}} > \frac{{15}}{{24}}\]

Vậy: \[\frac{5}{8} + \frac{3}{4}\,\, > \,\,\frac{1}{6} + \frac{{11}}{{24}}\]