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

Tính: a) 3 /5 + 6/5 = . . .

11/30

Tính:

a) \[\frac{3}{5} + \frac{6}{5} = ...\]          b) \[\frac{7}{{11}} + \frac{2}{{11}} = ...\]        c) \[\frac{{56}}{{31}} - \frac{{15}}{{31}} = ...\]                              d) \[\frac{{25}}{9} - \frac{8}{9} = ...\]

e) \[\frac{{15}}{{48}} + \frac{{21}}{{48}} = ...\]                         f) \[\frac{6}{{41}} + \frac{{15}}{{41}} = ...\]                            g) \[\frac{7}{{10}} - \frac{4}{{10}} = ...\]        h) \[\frac{{50}}{{27}} - \frac{{21}}{{27}} = ...\]

i) \[\frac{9}{{21}} + \frac{5}{{21}} = ...\]j) \[\frac{{12}}{{19}} + \frac{5}{{19}} = ...\]    k) \[\frac{{15}}{{16}} - \frac{9}{{16}} = ...\]l) \[\frac{{12}}{5} - \frac{1}{5} = ...\]

0/3000 ký tự
Giải thích

a) \[\frac{3}{5} + \frac{6}{5} = \frac{9}{5}\]                          b) \[\frac{7}{{11}} + \frac{2}{{11}} = \frac{9}{{11}}\]         c) \[\frac{{56}}{{31}} - \frac{{15}}{{31}} = \frac{{41}}{{31}}\]   d) \[\frac{{25}}{9} - \frac{8}{9} = \frac{{17}}{9}\]

e) \[\frac{{15}}{{48}} + \frac{{21}}{{48}} = \frac{{36}}{{48}}\] f) \[\frac{6}{{41}} + \frac{{15}}{{41}} = \frac{{21}}{{41}}\]   g) \[\frac{7}{{10}} - \frac{4}{{10}} = \frac{3}{{10}}\]    h) \[\frac{{50}}{{27}} - \frac{{21}}{{27}} = \frac{{29}}{{27}}\]

i) \[\frac{9}{{21}} + \frac{5}{{21}} = \frac{{14}}{{21}}\]       j) \[\frac{{12}}{{19}} + \frac{5}{{19}} = \frac{{17}}{{19}}\]    k) \[\frac{{15}}{{16}} - \frac{9}{{16}} = \frac{6}{{16}}\]    l) \[\frac{{12}}{5} - \frac{1}{5} = \frac{{11}}{5}\]