Ta có:
$A = \dfrac{100^{20} - 1}{100^{21} - 98}$
$100A = \dfrac{100^{21} - 100}{100^{21} - 98}$
$100A = \dfrac{100^{21} - 98 -2}{100^{21} - 98}$
$100A = 1 - \dfrac{2}{100^{21} - 98}$
$B = \dfrac{100^{19} - 1}{100^{20} - 98}$
$100B = \dfrac{100^{20} - 100}{100^{20} - 98}$
$100B = \dfrac{100^{20} - 98 -2}{100^{20} - 98}$
$100B = 1 - \dfrac{2}{100^{20} - 98}$
Vì : $\dfrac{2}{100^{21} - 98} < \dfrac{2}{100^{20}-98}$
$⇒ 100A > 100B$
$⇒ A > B$
