Ta có: \(A=\frac{97^{98}+1}{97^{99}+1}\Rightarrow97A=\frac{97^{99}+97}{97^{99}+1}=\frac{97^{99}+1+96}{97^{99}+1}=1+\frac{96}{97^{99}+1}\)
\(B=\frac{97^{97}+1}{97^{98}+1}\Rightarrow97B=\frac{97^{98}+97}{97^{98}+1}=\frac{97^{98}+1+96}{97^{98}+1}=1+\frac{96}{97^{98}+1}\)
Vì \(\frac{96}{97^{99}+1}< \frac{96}{97^{98}+1}\Rightarrow1+\frac{96}{97^{99}+1}< 1+\frac{96}{97^{98}+1}\Rightarrow97A< 97B\Rightarrow A< B\)
Vậy A < B
Ta thấy A < 1 và 96 > 1 nên ta có:
A < 9798 + 1 + 96 / 9799 + 1 + 96
=> A < 9798 + 97 / 9799 + 97
=> A < 97(9797 + 1) / 97(9798 + 1)
=> A < 9797 + 1 / 9798 + 1 = B
=> A < B
