Đáp án đúng: A
Giải chi tiết:\(A = \frac{{1 + \frac{1}{3} + \frac{1}{5} + \ldots + \frac{1}{{2017}} + \frac{1}{{2019}}}}{{\frac{1}{{1.2019}} + \frac{1}{{3.2017}} + \frac{1}{{5.2015}} + \ldots + \frac{1}{{2017.3}} + \frac{1}{{2019.1}}}}\)
Biến đổi tử số ta được:
\(\begin{array}{l}\,\,\,\,\,1 + \frac{1}{3} + \frac{1}{5} + \ldots + \frac{1}{{2017}} + \frac{1}{{2019}}\\ = \left( {1 + \frac{1}{{2019}}} \right) + \left( {\frac{1}{3} + \frac{1}{{2017}}} \right) + \ldots + \left( {\frac{1}{{1009}} + \frac{1}{{1011}}} \right)\\ = \left( {\frac{{2019}}{{1.2019}} + \frac{1}{{2019.1}}} \right) + \left( {\frac{{2017}}{{3.2017}} + \frac{3}{{3.2017}}} \right) + \ldots + \left( {\frac{{1011}}{{1009.1011}} + \frac{{1009}}{{1009.1011}}} \right)\\ = \frac{{2019 + 1}}{{1.2019}} + \frac{{2017 + 3}}{{3.2017}} + \ldots + \frac{{1011 + 1009}}{{1009.1011}}\\ = \frac{{2020}}{{1.2019}} + \frac{{2020}}{{3.2017}} + \ldots + \frac{{2020}}{{1009.1011}}\\ = 2020 \cdot \left( {\frac{1}{{1.2019}} + \frac{1}{{3.2017}} + \ldots + \frac{1}{{1009.1011}}} \right)\end{array}\)
\(\begin{array}{l} \Rightarrow A = \frac{{1 + \frac{1}{3} + \frac{1}{5} + \ldots + \frac{1}{{2017}} + \frac{1}{{2019}}}}{{\frac{1}{{1.2019}} + \frac{1}{{3.2017}} + \frac{1}{{5.2015}} + \ldots + \frac{1}{{2017.3}} + \frac{1}{{2019.1}}}}\\\,\,\,\,\,\,\,\,\,\,\, = \frac{{2020 \cdot \left( {\frac{1}{{1.2019}} + \frac{1}{{3.2017}} + \ldots + \frac{1}{{1009.1011}}} \right)}}{{2 \cdot \left( {\frac{1}{{1.2019}} + \frac{1}{{3.2017}} + \ldots + \frac{1}{{1009.1011}}} \right)}}\\\,\,\,\,\,\,\,\,\,\,\, = 1010.\end{array}\)
Vậy \(A = 1010.\)
Chọn A.