Điền dấu >, <, = thích hợp vào chỗ trống:

a) \(\frac{2}{5} + \frac{3}{{10}}\) = \(\frac{7}{{10}}\)

Giải thích

\(\frac{2}{5} + \frac{3}{{10}} = \frac{4}{{10}} + \frac{3}{{10}} = \frac{7}{{10}}\)

Vì \(\frac{7}{{10}}\) = \(\frac{7}{{10}}\) nên \(\frac{2}{5} + \frac{3}{{10}}\) = \(\frac{7}{{10}}\)

b) \(\frac{4}{9}\) < \(\frac{2}{9} + \frac{1}{3}\)

Giải thích

\(\frac{2}{9} + \frac{1}{3} = \frac{2}{9} + \frac{3}{9} = \frac{5}{9}\)

Vì \(\frac{4}{9}\) < \(\frac{5}{9}\) nên \(\frac{4}{9}\) < \(\frac{2}{9} + \frac{1}{3}\)

c) \(\frac{5}{8} + \frac{1}{4}\) > \(\frac{3}{4}\)

Giải thích

\(\frac{5}{8} + \frac{1}{4} = \frac{5}{8} + \frac{2}{8} = \frac{7}{8}\); \\(\frac{1}{3} + \frac{1}{4}\)(\frac{3}{4} = \frac{6}{8}\)

Vì \(\frac{7}{8} > \frac{6}{8}\) nên \(\frac{5}{8} + \frac{1}{4}\) > \(\frac{3}{4}\)

d) \(\frac{7}{{12}}\) = \(\frac{1}{3} + \frac{1}{4}\)

Giải thích

\(\frac{1}{3} + \frac{1}{4} = \frac{4}{{12}} + \frac{3}{{12}} = \frac{7}{{12}}\)

Vì \(\frac{7}{{12}}\) = \(\frac{7}{{12}}\) nên \(\frac{7}{{12}}\) =