Câu trả lời:
\(A=\frac{19^5-1+2016}{19^5-1}=1+\frac{2016}{19^5-1}\)
\(B=\frac{19^5-2+2016}{19^5-2}=1+\frac{2016}{19^5-2}\)
\(19^5-1>19^5-2\Rightarrow\frac{2016}{19^5-1}<\frac{2016}{19^5-2}\Rightarrow1+\frac{2016}{19^5-1}<1+\frac{2016}{19^5-2}\)
=> A<B