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Giải:
Ta có:
2019.2020-1/2019.2020= 2019.2020/2019.2020 - 1/2019.2020
=1-1/2019.2020
Tương tự:
2020.2021-1/2020.2021= 1-1/2020.2021
Vì 1/2019.2020 > 1/2020.2021 nên -1/2019.2020 < -1/2020.2021
(vì là số nguyên âm)
⇒ 1-1/2019.2020 < 1-1/2020.2021
⇔ 2019.2020-1/2019.2020 < 2020.2021-1/2020.2021
Chúc bạn học tốt!
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-Nghỉ Tết đi :)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)=\(1-\dfrac{1}{2022}\)=\(\dfrac{2021}{2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{2020}-\dfrac{1}{2021}=\dfrac{2021}{2020.2021}-\dfrac{2020}{2020.2021}=\dfrac{2021-2020}{2020.2021}=\dfrac{1}{2020.2021}\)
\(\dfrac{1}{2020\cdot2021}=\dfrac{2021-2020}{2020\cdot2021}=\dfrac{1}{2020}-\dfrac{1}{2021}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
S=1/2x3+1/4x5+1/6x7+...+1/2022x2023<1/2x3+1/3x4+1/4x5+...+1/1010x1011
=1/2-1/1011=1009/2022<1011/2023
=>S<1011/2023
S= 1/2.3 + 1/4.5 + 1/6.7 +.....+ 1 2020.2021 + 1 2022.2023 . : So sánh S và 1011/2023
![](https://rs.olm.vn/images/avt/0.png?1311)
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a. 67/77 = 1 - 10/77; 73/83=1 - 10/83
Vì 10/77>10/83 nên 1 - 10/77 < 1-10/83
Vậy 67/77<73/83
c. Ta có: n/n+3 < n+1/n+3 <n+1/n+2
Vậy n/n+3 < n+1/n+2
![](https://rs.olm.vn/images/avt/0.png?1311)
C = 3(4/3.7 + 4/7.11 + ... + 4/2021.2022)
= 3(1/3 - 1/7 + 1/7 - 1/11 + ... + 1/2021 - 1/2022)
= 3(1/3 - 1/2022)
= 3(2022-3)/3.2022
= 2019 / 2022