\(a,\dfrac{1}{7}và\dfrac{4}{5}\)
\(\dfrac{1}{7}=\dfrac{1\times5}{7\times5}=\dfrac{5}{35}\)
\(\dfrac{4}{5}=\dfrac{4\times7}{5\times7}=\dfrac{28}{35}\)
Vậy \(\dfrac{5}{35}< \dfrac{28}{35}hay\dfrac{1}{7}< \dfrac{4}{5}\)
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