Ta có: \(A=\dfrac{19^5+2016}{19^5-1}=1+\dfrac{2017}{19^5-1}\)
\(B=\dfrac{19^5+2015}{19^5-2}=1+\dfrac{2017}{19^5-2}\)
Vì \(\dfrac{2017}{19^5-1}< \dfrac{2017}{19^5-2}\Rightarrow1+\dfrac{2017}{19^5-1}< 1+\dfrac{2017}{19^5-2}\)
\(\Rightarrow A< B\)
Vậy A < B