Đáp án đúng: A
Giải chi tiết:\(A = \frac{{{{10}^{2019}} + 1}}{{{{10}^{2020}} + 1}}\) và \(B = \frac{{{{10}^{2020}} + 1}}{{{{10}^{2021}} + 1}}\).
Ta có: \(A = \frac{{{{10}^{2019}} + 1}}{{{{10}^{2020}} + 1}}\)
\( \Rightarrow 10.A = 10 \cdot \frac{{{{10}^{2019}} + 1}}{{{{10}^{2020}} + 1}} = \frac{{{{10}^{2020}} + 10}}{{{{10}^{2020}} + 1}} = \frac{{{{10}^{2020}} + 1 + 9}}{{{{10}^{2020}} + 1}}\)
\( = \frac{{{{10}^{2020}} + 1}}{{{{10}^{2020}} + 1}} + \frac{9}{{{{10}^{2020}} + 1}} = 1 + \frac{9}{{{{10}^{2020}} + 1}}\)
\(\begin{array}{l}B = \frac{{{{10}^{2020}} + 1}}{{{{10}^{2021}} + 1}} \Rightarrow 10.B = 10 \cdot \frac{{\left( {{{10}^{2020}} + 1} \right)}}{{{{10}^{2021}} + 1}}\\ \Rightarrow 10.B = \frac{{10.\left( {{{10}^{2020}} + 1} \right)}}{{{{10}^{2021}} + 1}} = \frac{{{{10}^{2021}} + 10}}{{{{10}^{2021}} + 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{10}^{2021}} + 1 + 9}}{{{{10}^{2021}} + 1}} = \frac{{{{10}^{2021}} + 1}}{{{{10}^{2021}} + 1}} + \frac{9}{{{{10}^{2021}} + 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \frac{9}{{{{10}^{2021}} + 1}}.\end{array}\)
Vì \({10^{2020}} < {10^{2021}} \Rightarrow {10^{2020}} + 1 < {10^{2021}} + 1 \Rightarrow \frac{1}{{{{10}^{2020}} + 1}} > \frac{1}{{{{10}^{2021}} + 1}}\)
\( \Rightarrow \frac{9}{{{{10}^{2020}} + 1}} > \frac{9}{{{{10}^{2021}} + 1}}\)
\( \Rightarrow 1 + \frac{9}{{{{10}^{2020}} + 1}} > 1 + \frac{9}{{{{10}^{2021}} + 1}}\)
\( \Rightarrow 10A > 10B \Rightarrow A > B\)
Chọn A.
