So sánh: A=( 98^99 +10/( 98^89 +1) và B= ( 98^98 +1)/(98^88 +1)

* Hướng dẫn giải

a) Do $A=\frac{{98}^{99}+1}{{98}^{89}+1}>1$ nên

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

Vậy A > B

b) Do C = $\frac{{100}^{2008}+1}{{100}^{2018}+1}$ < 1 nên

C= $\frac{{100}^{2008}+1}{{100}^{2018}+1}>\frac{{100}^{2008}+1+99}{{100}^{2018}+1+99}=\frac{100({100}^{2007}+1)}{100({100}^{2017}+1)}=\frac{{100}^{2007}+1}{{100}^{2017}+1}=D$

Vậy C > D.