Cho M = 2018/2019 + 2019/2020 + 2020/2021 + 2021/2018; M = 1/8 + 1/9 + 1/10
Câu hỏi:
Trả lời:
\(M = \frac{{2018}}{{2019}} + \frac{{2019}}{{2020}} + \frac{{2020}}{{2021}} + \frac{{2021}}{{2018}}\)
\(M = 1 - \frac{1}{{2019}} + 1 - \frac{1}{{2020}} + 1 - \frac{1}{{2021}} + 1 + \frac{3}{{2018}}\)
Do \(\frac{1}{{2019}} < \frac{1}{{2018}};\frac{1}{{2020}} < \frac{1}{{2018}};\frac{1}{{2021}} < \frac{1}{{2018}}\)
Suy ra: \(\frac{1}{{2019}} + \frac{1}{{2020}} + \frac{1}{{2021}} < \frac{3}{{2018}}\)
Suy ra: M > 3 (1)
Lại có: \(N = \frac{1}{8} + \frac{1}{9} + \frac{1}{{10}} + ... + \frac{1}{{62}} + \frac{1}{{63}}\)
Xét \(\frac{1}{8} + \frac{1}{9} + \frac{1}{{10}} + ... + \frac{1}{{15}} < \frac{1}{8}.8 = 1\)
\(\frac{1}{{16}} + \frac{1}{{17}} + \frac{1}{{18}} + ... + \frac{1}{{31}} < \frac{1}{{16}}.16 = 1\)
\(\frac{1}{{32}} + \frac{1}{{33}} + \frac{1}{{34}} + ... + \frac{1}{{63}} < \frac{1}{{32}}.32 = 1\)
Suy ra: N < 1 + 1 + 1 = 3 (2)
Từ (1) và (2) suy ra: M > N.
