Cho hai biểu thức: Cho hai biểu thức: A = 1/ 2 + 1/ 3 + 1/ 4 + . . . . + 1 /2020 + 1 /2021 + 1 /2022 ; B = 2021 /1 + 2020 /2 + 2019 /3 + . . . + 2 /2020 + 1 /2021 . Tính: A/ B .

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(1,0 điểm)Cho hai biểu thức:

\(A = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + .... + \frac{1}{{2020}} + \frac{1}{{2021}} + \frac{1}{{2022}}\);

\(B = \frac{{2021}}{1} + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{2}{{2020}} + \frac{1}{{2021}}\).

Tính: \(\frac{A}{B}\).

Nhập câu trả lời ở đây

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Giải thích

Ta có \(B = \frac{{2021}}{1} + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{2}{{2020}} + \frac{1}{{2021}}\)

\( = 1 + \left( {\frac{{2020}}{2} + 1} \right) + \left( {\frac{{2019}}{3} + 1} \right) + \left( {\frac{{2018}}{4} + 1} \right) + ... + \left( {\frac{2}{{2020}} + 1} \right) + \left( {\frac{1}{{2021}} + 1} \right)\)

\( = 1 + \frac{{2020 + 2}}{2} + \frac{{2019 + 3}}{3} + ... + \frac{{2 + 2020}}{{2020}} + \frac{{1 + 2021}}{{2021}}\)

\( = \frac{{2022}}{{2022}} + \frac{{2022}}{2} + \frac{{2022}}{3} + ... + \frac{{2022}}{{2022}} + \frac{{2022}}{{2021}}\)

\( = 2022\left( {\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2020}} + \frac{1}{{2021}} + \frac{1}{{2022}}} \right) = 2022A\).

Vậy \(\frac{A}{B} = \frac{A}{{2022A}} = \frac{1}{{2022}}\).