So sánh: 

- Hướng dẫn giải

Do $A=\frac{{98}^{99}+1}{{98}^{89}+1}>1$ nên $A=\frac{{98}^{99}+1}{{98}^{89}+1}>\frac{{98}^{99}+1+97}{{98}^{89}+1+97}=\frac{98\left({98}^{98}+1\right)}{98\left({98}^{88}+1\right)}=\frac{{98}^{98}+1}{{98}^{88}+1}=B$

Vậy A > B