Ta có:
\(A=\frac{17^{18}+1}{17^{19}+1}\)
\(\Rightarrow17A=\frac{17^{19}+1+16}{17^{19}+1}\)
\(\Rightarrow17A=1+\frac{16}{17^{19}+1}\)
\(B=\frac{17^{17}+1}{17^{18}+1}\)
\(\Rightarrow17B=\frac{17^{18}+1+16}{17^{18}+1}\)
\(\Rightarrow17B=1+\frac{16}{17^{18}+1}\)
Vì \(\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\Rightarrow17A< 17B\)
\(\Rightarrow A< B\)
Vậy \(A< B\)