Điền dấu thích hợp vào chỗ chấm:
\(a)\,\frac{3}{4}-\frac{1}{2}\,\cdot \cdot \cdot \,\frac{1}{6}\,\)
\(b)\,\,\,\frac{3}{8}+\frac{2}{5}\,\,\,\cdot \cdot \cdot \,\,\,\frac{1}{4}+\frac{9}{20}\)
\(c)\,\,\frac{5}{4}+\frac{1}{2}-1\,\cdot \cdot \cdot \,\frac{3}{4}-\frac{2}{15}+\frac{9}{30}\,\)
Câu 286222: Điền dấu thích hợp vào chỗ chấm:
\(a)\,\frac{3}{4}-\frac{1}{2}\,\cdot \cdot \cdot \,\frac{1}{6}\,\)
\(b)\,\,\,\frac{3}{8}+\frac{2}{5}\,\,\,\cdot \cdot \cdot \,\,\,\frac{1}{4}+\frac{9}{20}\)
\(c)\,\,\frac{5}{4}+\frac{1}{2}-1\,\cdot \cdot \cdot \,\frac{3}{4}-\frac{2}{15}+\frac{9}{30}\,\)
A. a) \(\frac{3}{4} - \frac{1}{2} > \frac{1}{6}\)
b) \(\frac{3}{8} + \frac{2}{5}\,\,\, > \,\,\,\frac{1}{4} + \frac{9}{{20}}\)
c) \(\frac{5}{4} + \frac{1}{2} - 1\,\,\,\, > \,\,\,\,\frac{3}{4} - \frac{2}{{15}} + \frac{9}{{30}}\,\)
B. a) \(\frac{3}{4} - \frac{1}{2} > \frac{1}{6}\)
b) \(\frac{3}{8} + \frac{2}{5}\,\,\, < \,\,\,\frac{1}{4} + \frac{9}{{20}}\)
c) \(\frac{5}{4} + \frac{1}{2} - 1\,\,\,\, < \,\,\,\,\frac{3}{4} - \frac{2}{{15}} + \frac{9}{{30}}\,\)
C. a) \(\frac{3}{4} - \frac{1}{2} > \frac{1}{6}\)
b) \(\frac{3}{8} + \frac{2}{5}\,\,\, > \,\,\,\frac{1}{4} + \frac{9}{{20}}\)
c) \(\frac{5}{4} + \frac{1}{2} - 1\,\,\,\, < \,\,\,\,\frac{3}{4} - \frac{2}{{15}} + \frac{9}{{30}}\,\)
D. a) \(\frac{3}{4} - \frac{1}{2} < \frac{1}{6}\)
b) \(\frac{3}{8} + \frac{2}{5}\,\,\, > \,\,\,\frac{1}{4} + \frac{9}{{20}}\)
c) \(\frac{5}{4} + \frac{1}{2} - 1\,\,\,\, < \,\,\,\,\frac{3}{4} - \frac{2}{{15}} + \frac{9}{{30}}\,\)
Thực hiện các phép tính ở hai vế rồi so sánh kết quả của các phép tính đó.
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Đáp án : C(0) bình luận (0) lời giải
Giải chi tiết:
a) Ta có: \(\frac{3}{4} - \frac{1}{2}\, = \frac{3}{4} - \frac{2}{4} = \frac{1}{4}\)
Mà \(\frac{1}{4} > \frac{1}{6}\), do đó \(\frac{3}{4} - \frac{1}{2} > \frac{1}{6}\)b) \(\,\,\frac{3}{8} + \frac{2}{5}\, = \frac{{15}}{{40}} + \frac{{16}}{{40}} = \frac{{31}}{{40}}\)
\(\frac{1}{4} + \frac{9}{{20}} = \frac{5}{{20}} + \frac{9}{{20}} = \frac{{14}}{{20}} = \frac{{28}}{{40}}\)
Mà \(\frac{{31}}{{40}} > \frac{{28}}{{40}}\), do đó \(\frac{3}{8} + \frac{2}{5}\,\,\, > \,\,\,\frac{1}{4} + \frac{9}{{20}}\).
\(\begin{array}{l}c)\;\;\frac{5}{4} + \frac{1}{2} - 1\, = \frac{5}{4} + \frac{2}{4} - 1 = \frac{7}{4} - 1 = \frac{7}{4} - \frac{4}{4} = \frac{3}{4};\\\;\;\;\;\frac{3}{4} - \frac{2}{{15}} + \frac{9}{{30}} = \frac{{45}}{{60}} - \frac{8}{{60}} + \frac{3}{{10}} = \frac{{37}}{{60}} + \frac{3}{{10}} = \frac{{37}}{{60}} + \frac{{18}}{{60}} = \frac{{55}}{{60}} = \frac{{11}}{{12}}\,\end{array}\)
Ta có: \(\frac{3}{4} = \frac{9}{{12}}\) . Mà \(\frac{9}{{12}} < \frac{{11}}{{12}}\)
Do đó \(\frac{5}{4} + \frac{1}{2} - 1\,\,\,\, < \,\,\,\,\frac{3}{4} - \frac{2}{{15}} + \frac{9}{{30}}\,\)
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