Bộ 20 đề thi cuối kì 2 Toán lớp 4 Cánh diều có đáp án - Đề 11

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\[\frac{7}{9}\,\, + \,\,\frac{4}{3}\,\,.............\,\,\frac{8}{3}\,\, - \,\,1\]

\[\frac{5}{9}\,\, - \,\,\frac{1}{3}\,\,...............\,\,\frac{2}{9}\]

\[\frac{7}{9}\,\, \times \,\,\frac{3}{7}\,\,..............\,\,\frac{1}{2}\]

\[\frac{4}{9}\,\, + \,\,\frac{7}{3}\,\,.............\,\,2\]

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Giải thích

\[\frac{7}{9}\,\, + \,\,\frac{4}{3}\,\,\,\, > \,\,\,\,\frac{8}{3}\,\, - \,\,1\]

Giải thích:

\[\frac{7}{9}\,\, + \,\,\frac{4}{3}\, = \frac{7}{9}\,\, + \,\,\frac{{12}}{9} = \frac{{19}}{9}\]

\[\frac{8}{3}\,\, - \,\,1 = \frac{8}{3}\,\, - \,\,\frac{3}{3} = \frac{5}{3} = \frac{{15}}{9}\]

Do \[\frac{{19}}{9}\]> \[\frac{{15}}{9}\] nên \[\frac{7}{9}\,\, + \,\,\frac{4}{3}\,\,\,\, > \,\,\,\,\frac{8}{3}\,\, - \,\,1\]

 

\[\frac{5}{9}\,\, - \,\,\frac{1}{3}\,\,\,\, = \,\,\,\,\frac{2}{9}\]

Giải thích:

\[\frac{5}{9}\,\, - \,\,\frac{1}{3} = \,\frac{5}{9}\,\, - \,\,\frac{3}{9}\, = \,\,\,\,\frac{2}{9}\]

\[\frac{7}{9}\,\, \times \,\,\frac{3}{7}\,\,\,\, < \,\,\,\,\frac{1}{2}\]

Giải thích:

Do \[\frac{7}{9}\,\, \times \,\,\frac{3}{7}\, = \frac{1}{3}\]\[\, < \,\frac{1}{2}\]

\[\frac{4}{9}\,\, + \,\,\frac{7}{3}\,\,\,\, > \,\,\,2\]

Giải thích:

\[\frac{4}{9}\,\, + \,\,\frac{7}{3}\,\, = \frac{4}{9}\,\, + \frac{{21}}{9} = \frac{{25}}{9}\];   \(2 = \frac{{18}}{9}\)

Do \[\frac{{25}}{9}\]> \(\frac{{18}}{9}\) nên \[\frac{4}{9}\,\, + \,\,\frac{7}{3}\,\,\,\, > \,\,\,2\]