Question

[Phùng Thị Như Quỳnh]

so sánh^{\(\frac{2017^6+5}{2017^6+3}va\frac{2017^6+7}{2017^6+5}\)}

Answer 1

[Eihwaz]

\(\frac{2017^6+5}{2017^6+3}=1+\frac{2}{2017^6+3}>1+\frac{2}{2017^6+5}.\)=\(\frac{2017^6+7}{2017^6+5}\)

Answer 2

[Phùng Thị Như Quỳnh]

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