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

\[\frac{7}{{12}}\,\, + \,\,2\,\,\,\,\, < \,\,\,\,\,3\,\, - \,\,\frac{1}{3}\]

Giải thích:

\[\frac{7}{{12}}\,\, + \,\,2 = \frac{7}{{12}}\,\, + \,\,\frac{{24}}{{12}} = \frac{{31}}{{12}}\,\,\]

\[3\,\, - \,\,\frac{1}{3} = \frac{9}{3} - \,\,\frac{1}{3} = \frac{8}{3} = \frac{{32}}{{12}}\]

Do \[\frac{{31}}{{12}}\,\]< \[\frac{{32}}{{12}}\] nên \[\frac{7}{{12}}\,\, + \,\,2\,\,\,\,\, < \,\,\,\,\,3\,\, - \,\,\frac{1}{3}\]

\[\frac{4}{9}\,\, \times \,\,\frac{3}{7}\,\,\,\,\, < \,\,\,\,1\]

Giải thích:

\[\frac{4}{9}\,\, \times \,\,\frac{3}{7}\, = \frac{4}{{21}} < 1\]

\[\frac{5}{{11}}\,\,:\,\,\frac{3}{{11}}\,\,\,\, < \,\,\,\,2\]

Giải thích:

\[\frac{5}{{11}}\,\,:\,\,\frac{3}{{11}}\, = \,\frac{5}{{11}} \times \frac{{11}}{3} = \frac{5}{3}\]

\(2 = \frac{6}{3}\)

Do \[\frac{5}{3}\] < \(\frac{6}{3}\) nên \[\frac{5}{{11}}\,\,:\,\,\frac{3}{{11}}\,\,\,\, < \,\,\,\,2\]

\[5\,\, - \,\,\frac{4}{9}\,\,\,\,\, > \,\,\,\,\frac{5}{3}\,\, + \,\,\frac{5}{9}\]

Giải thích:

\[5\,\, - \,\,\frac{4}{9} = \frac{{45}}{9} - \,\,\frac{4}{9} = \,\frac{{41}}{9}\]

\[\frac{5}{3}\,\, + \,\,\frac{5}{9} = \,\frac{{15}}{9} + \,\,\frac{5}{9} = \frac{{20}}{9}\]

Do \[\frac{{41}}{9}\]> \[\frac{{20}}{9}\] nên \[5\,\, - \,\,\frac{4}{9}\,\,\,\,\, > \,\,\,\,\frac{5}{3}\,\, + \,\,\frac{5}{9}\]