Đáp án:
$A > B$
Giải thích các bước giải:
Ta có:
$+)\quad 1+ 19+ 19^2 +\dots + 19^{2020}$
$= \dfrac{19^{2021} -1}{18}$
$\to A =\dfrac{19^{2021}}{1+ 19+ 19^2 +\dots + 19^{2020}}$
$\to A =\dfrac{18.19^{2021}}{19^{2021} -1}$
$+)\quad 1+ 18+ 18^2 +\dots + 18^{2020}$
$=\dfrac{18^{2021} -1}{17}$
$\to B =\dfrac{17.18^{2021}}{18^{2021}-1}$
Xét $A - B =\dfrac{18.19^{2021}}{19^{2021} -1} - \dfrac{17.18^{2021}}{18^{2021} -1}$
$=\dfrac{18.19^{2021}(18^{2021} - 1) - 17.18^{2021}(19^{2021} -1)}{(19^{2021} -1)(18^{2021} -1)}$
Ta có:
$\quad 18.19^{2021}(18^{2021} - 1) - 17.18^{2021}(19^{2021} -1)$
$= 18.(18.19)^{2021} - 18.19^{2021} - 17.(18.19)^{2021} + 17.18^{2021}$
$= (18.19)^{2021} + 17.18^{2021} - 18.19^{2021} > 0$
Lại có: $(19^{2021} -1)(18^{2021} -1)>0$
$\to \dfrac{18.19^{2021}(18^{2021} - 1) - 17.18^{2021}(19^{2021} -1)}{(19^{2021} -1)(18^{2021} -1)}$
$\to A - B > 0$
$\to A > B$
