Đáp án:
Ta có :
A = $\frac{10^{99}+1}{10^{98}+1}$
=> $\frac{A}{10}$ = $\frac{10^{99}+1}{10^{99}+10}$ = 1 - $\frac{9}{10^{99}+10}$
B = $\frac{10^{98}+1}{10^{97}+1}$
=> $\frac{B}{10}$ = $\frac{10^{98}+1}{10^{98}+10}$ = 1 - $\frac{9}{10^{98}+10}$
Do $10^{99} > 10^{98}$
$=> 10^{99} + 10 > 10^{98} + 10$
$=> \frac{9}{10^{99}+10} < \frac{9}{10^{98}+10}$
=> $1 - \frac{9}{10^{99}+10} > 1- \frac{9}{10^{98}+10}$
=> $\frac{A}{10}$ > $\frac{B}{10}$
Giải thích các bước giải:
