Đề cương ôn tập giữa kì 2 Toán lớp 4 Cánh diều có đáp án - Phần II. Tự luận

Điền dấu >, <, = thích hợp vào chỗ chấm a) 7/12 + 15/12 . . . 1/6 + 7/6

13/30

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

a) \[\frac{7}{{12}} + \frac{{15}}{{12}}\,\,...\,\,\frac{1}{6} + \frac{7}{6}\]               b) \[\frac{{16}}{{15}} + \frac{{14}}{{15}}\,\,...\,\,\frac{3}{5} + \frac{7}{5}\]

c) \[\frac{{19}}{{15}} - \frac{4}{{15}}\,\,...\,\,\frac{7}{5} - \frac{3}{5}\]                d) \[\frac{{19}}{{16}} - \frac{7}{{16}}\,\,...\,\,\frac{9}{4} - \frac{3}{4}\]

e) \[\frac{8}{9} + \frac{5}{9}\,\,...\,\,\frac{5}{3} - \frac{1}{3}\]           f) \[\frac{{19}}{{10}} - \frac{7}{{10}}\,\,...\,\,\frac{6}{5} + \frac{7}{5}\]

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Giải thích

a)

\[\frac{7}{{12}} + \frac{{15}}{{12}} = \frac{{22}}{{12}} = \frac{{11}}{6}\]

\[\frac{1}{6} + \frac{7}{6} = \frac{8}{6}\]

So sánh: \[\frac{{11}}{6} > \frac{8}{6}\]

Vậy: \[\frac{7}{{12}} + \frac{{15}}{{12}}\,\, > \,\,\frac{1}{6} + \frac{7}{6}\]

b)

\[\frac{{16}}{{15}} + \frac{{14}}{{15}} = \frac{{30}}{{15}} = 2\]

\[\frac{3}{5} + \frac{7}{5} = \frac{{10}}{5} = 2\]

So sánh: 2 = 2

Vậy: \[\frac{{16}}{{15}} + \frac{{14}}{{15}}\,\, = \,\,\frac{3}{5} + \frac{7}{5}\]

c)

\[\frac{{19}}{{15}} - \frac{4}{{15}} = \frac{{15}}{{15}} = 1\]

\[\frac{7}{5} - \frac{3}{5} = \frac{4}{5}\]

So sánh: \[\frac{4}{5} < 1\]

Vậy: \[\frac{{19}}{{15}} - \frac{4}{{15}}\,\, > \,\,\frac{7}{5} - \frac{3}{5}\]

d)

\[\frac{{19}}{{16}} - \frac{7}{{16}} = \frac{{12}}{{16}} = \frac{3}{4}\]

\[\frac{9}{4} - \frac{3}{4} = \frac{6}{4}\]

So sánh: \[\frac{3}{4} < \frac{6}{4}\]

Vậy: \[\frac{{19}}{{16}} - \frac{7}{{16}}\,\, < \,\,\frac{9}{4} - \frac{3}{4}\]

e)

\[\frac{8}{9} + \frac{5}{9}\, = \frac{{13}}{9}\]

\[\frac{5}{3} - \frac{1}{3} = \frac{4}{3} = \frac{{12}}{9}\]

So sánh: \[\frac{{13}}{9} > \frac{{12}}{9}\]

Vậy: \[\frac{8}{9} + \frac{5}{9}\,\, > \,\,\frac{5}{3} - \frac{1}{3}\]

f)

\[\frac{{19}}{{10}} - \frac{7}{{10}}\, = \frac{{12}}{{10}} = \frac{6}{5}\]

\[\frac{6}{5} + \frac{7}{5} = \frac{{13}}{5}\]

So sánh: \[\frac{6}{5} < \frac{{13}}{5}\]

Vậy: \[\frac{{19}}{{10}} - \frac{7}{{10}}\,\, < \,\,\frac{6}{5} + \frac{7}{5}\]