Câu hỏi:

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Tự luận

Cho hai biểu thức:

A=12+13+14+...+12022A = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2022}} B=20211+20202+20193+...+12021B = \frac{{2021}}{1} + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{1}{{2021}}.

Tính tỉ số BA\frac{B}{A}.

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Lời giải

Hướng dẫn giải:

Ta có: A=12+13+14+...+12022A = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2022}};

B=20211+20202+20193+...+12021B = \frac{{2021}}{1} + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{1}{{2021}}

 =2021+20202+20193+...+12021 = 2021 + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{1}{{2021}}

=2020+20202+20193+...+12021+1 = 2020 + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{1}{{2021}} + 1

=(1+20202)+(1+20193)+...+(12021+1)+1 = \,\left( {1 + \frac{{2020}}{2}} \right) + \left( {1 + \frac{{2019}}{3}} \right) + ... + \left( {\frac{1}{{2021}} + 1} \right) + 1

=20222+20223+...+20222021+20222022 = \,\frac{{2022}}{2} + \frac{{2022}}{3} + ... + \frac{{2022}}{{2021}} + \frac{{2022}}{{2022}}

=2022  .  (12+13+...+12021+12022) = 2022\,\,.\,\,\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}} + \frac{1}{{2022}}} \right).

Khi đó,  BA=2022  .  (12+13+...+12021+12022)12+13+14+...+12022=2022\frac{B}{A} = \frac{{2022\,\,.\,\,\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}} + \frac{1}{{2022}}} \right)}}{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2022}}}} = 2022.

Vậy tỉ số BA=2022\frac{B}{A} = 2022.

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