Đề cương ôn tập giữa kì 2 Toán lớp 4 Cánh diều có đáp án - Phần II. Tự luận
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Điền dấu >, <, = thích hợp vào chỗ chấm a) 5/3 + 1/6 . . . 11/6

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

a) \[\frac{5}{3} + \frac{1}{6}\,\,...\,\,\frac{{11}}{6}\]                  b) \[\frac{5}{{12}} + \frac{2}{3}\,\,...\,\,\frac{{11}}{{12}}\]      c) \[\frac{9}{5} + \frac{3}{{10}}\,\,...\,\,\frac{{21}}{{10}}\]  d) \[\frac{{14}}{{15}}\,\,...\,\,\frac{7}{{15}} + \frac{1}{{15}}\]

e) \[\frac{6}{5} + \frac{7}{{10}}\,\,...\,\,\frac{7}{4} + \frac{{11}}{{20}}\]               f) \[\frac{3}{4} + \frac{7}{{12}}\,\,...\,\,\frac{1}{3} + \frac{5}{{12}}\]

g) \[\frac{7}{8} + \frac{3}{2}\,\,...\,\,\frac{5}{4} + \frac{9}{{16}}\]              h) \[\frac{5}{2} + \frac{7}{4}\,\,...\,\,\frac{3}{2} + \frac{5}{8}\]

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a)

\[\frac{5}{3} + \frac{1}{6} = \frac{{10}}{6} + \frac{1}{6} = \frac{{11}}{6}\]

So sánh: \[\frac{{11}}{6} = \,\frac{{11}}{6}\]

Vậy: \[\frac{5}{3} + \frac{1}{6}\,\, = \,\,\frac{{11}}{6}\]       

b)

\[\frac{5}{{12}} + \frac{2}{3} = \frac{5}{{12}} + \frac{8}{{12}} = \frac{{13}}{{12}}\]

So sánh: \[\frac{{13}}{{12}}\,\, > \,\,\frac{{11}}{{12}}\]

Vậy: \[\frac{5}{{12}} + \frac{2}{3}\,\, > \,\,\frac{{11}}{{12}}\]  

c)

\[\frac{9}{5} + \frac{3}{{10}} = \frac{{18}}{{10}} + \frac{3}{{10}} = \frac{{21}}{{10}}\]

So sánh: \[\frac{{21}}{{10}}\,\, = \,\,\frac{{21}}{{10}}\]

Vậy: \[\frac{9}{5} + \frac{3}{{10}} = \frac{{21}}{{10}}\]

d)

 

\[\frac{7}{{15}} + \frac{1}{{15}} = \frac{8}{{15}}\]

So sánh: \[\frac{{14}}{{15}}\,\, > \,\,\frac{8}{{15}}\]

Vậy: \[\frac{{14}}{{15}}\,\, > \,\,\frac{7}{{15}} + \frac{1}{{15}}\]

e)

\[\frac{6}{5} + \frac{7}{{10}} = \frac{{12}}{{10}} + \frac{7}{{10}} = \frac{{19}}{{10}} = \frac{{38}}{{20}}\]

\[\frac{7}{4} + \frac{{11}}{{20}} = \frac{{35}}{{20}} + \frac{{11}}{{20}} = \frac{{46}}{{20}}\]

So sánh: \[\frac{{38}}{{20}} < \frac{{46}}{{20}}\]

Vậy: \[\frac{6}{5} + \frac{7}{{10}}\,\, < \,\,\frac{7}{4} + \frac{{11}}{{20}}\]

f)

\[\frac{3}{4} + \frac{7}{{12}} = \frac{9}{{12}} + \frac{7}{{12}} = \frac{{16}}{{12}}\]

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

So sánh: \[\frac{{16}}{{12}} > \frac{9}{{12}}\]

Vậy: \[\frac{3}{4} + \frac{7}{{12}}\,\, > \,\,\frac{1}{3} + \frac{5}{{12}}\]

g)

\[\frac{7}{8} + \frac{3}{2} = \frac{7}{8} + \frac{{12}}{8} = \frac{{19}}{8} = \frac{{38}}{{16}}\]

\[\frac{5}{4} + \frac{9}{{16}} = \frac{{20}}{{16}} + \frac{9}{{16}} = \frac{{29}}{{16}}\]

So sánh: \[\frac{{38}}{{16}} > \frac{{29}}{{16}}\]

Vậy: \[\frac{7}{8} + \frac{3}{2} > \,\,\frac{5}{4} + \frac{9}{{16}}\]

h)

\[\frac{5}{2} + \frac{7}{4} = \frac{{10}}{4} + \frac{7}{4} = \frac{{17}}{4} = \frac{{34}}{8}\]

\[\frac{3}{2} + \frac{5}{8} = \frac{{12}}{8} + \frac{5}{8} = \frac{{17}}{8}\]

So sánh: \[\frac{{34}}{8} > \frac{{17}}{8}\]

Vậy: \[\frac{5}{2} + \frac{7}{4}\,\, > \,\,\frac{3}{2} + \frac{5}{8}\]