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

\(\frac{4}{5}\,\, + \,\,\frac{1}{{10}}\) ……… \(\frac{1}{2}\)	\(\frac{5}{8}\,\, \times \,\,\frac{2}{5}\) ……… \(\frac{3}{4}\)
\(\frac{1}{2}\,\, - \,\,\frac{1}{4}\) ……… \(\frac{1}{4}\)	\(\frac{3}{4}\,\,:\,\,\frac{6}{9}\,\, - \,\,\frac{1}{2}\) ……… \(\frac{7}{8}\)

\(\frac{4}{5}\,\, + \,\,\frac{1}{{10}}\) >  \(\frac{1}{2}\)                                                  \(\frac{5}{8}\,\, \times \,\,\frac{2}{5}\) < \(\frac{3}{4}\)

\(\frac{1}{2}\,\, - \,\,\frac{1}{4}\) =  \(\frac{1}{4}\)                                                    \(\frac{3}{4}\,\,:\,\,\frac{6}{9}\,\, - \,\,\frac{1}{2}\) <  \(\frac{7}{8}\)

Giải thích:

- Ta có: \(\frac{4}{5} + \frac{1}{{10}} = \frac{8}{{10}} + \frac{1}{{10}} = \frac{9}{{10}}\); \(\frac{1}{2}\,\, = \,\,\frac{5}{{10}}\)                                                  

Vì \(\frac{9}{{10}}\,\, > \,\,\frac{5}{{10}}\) nên \(\frac{4}{5}\,\, + \,\,\frac{1}{{10}}\) >  \(\frac{1}{2}\)                                                                                                  

- Ta có: \[\frac{5}{8}\,\, \times \,\,\frac{2}{5}\,\, = \,\,\frac{{10}}{{40}}\,\, = \,\,\frac{1}{4}\]

Vì: \[\frac{1}{4} < \frac{3}{4}\] nên \(\frac{5}{8}\,\, \times \,\,\frac{2}{5}\) < \(\frac{3}{4}\)

- Ta có: \[\frac{1}{2}\,\, - \,\,\frac{1}{4}\,\, = \,\,\frac{2}{4}\,\, - \,\,\frac{1}{4}\,\, = \,\,\frac{1}{4}\]                                                         

Vì \[\frac{1}{4} = \frac{1}{4}\] nên \(\frac{1}{2}\,\, - \,\,\frac{1}{4}\) =  \(\frac{1}{4}\)                  

- Ta có: \[\frac{3}{4}\,\,:\,\,\frac{6}{9}\,\, - \,\,\frac{1}{2}\,\, = \,\,\frac{3}{4}\,\, \times \,\,\frac{9}{6}\,\, = \,\,\frac{9}{8}\,\, - \,\,\frac{1}{2}\,\, = \,\,\frac{9}{8}\,\, - \,\,\frac{4}{8}\,\, = \,\,\frac{5}{8}\]

Ví \(\frac{5}{8}\,\, < \,\,\frac{7}{8}\) nên \(\frac{3}{4}\,\,:\,\,\frac{6}{9}\,\, - \,\,\frac{1}{2}\) <  \(\frac{7}{8}\)