So sánh \(\dfrac{2019.2020-1}{2019.2020}\) và \(\dfrac{2020.2021-1}{2020.2021}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}+\dfrac{1}{2020\cdot2021}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2021}\)
\(=\dfrac{1}{1}-\dfrac{1}{2021}=\dfrac{2021}{2021}-\dfrac{1}{2021}\)
\(=\dfrac{2020}{2021}\)
mà \(\dfrac{2020}{2021}< \dfrac{2021}{2021}=1\)
nên A<1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)
\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)
=>B<1
=>A>B
![](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)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\\ =\dfrac{1}{1}-\dfrac{1}{2020}=\dfrac{2019}{2020}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
1/1x2+1/2x3+1/3x4+...+1/2020x2021+1/2021x2022
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021+1/2021-1/2022.
=1/1-1/2022
=2021/2022
![](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)
\(A=\frac{1.2}{2.2}\cdot\frac{2.3}{3.3}\cdot\frac{3.4}{4.4}\cdot...\cdot\frac{2020.2021}{2021.2021}\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2020}{2021}\)
\(A=\frac{1.2.3.....2020}{2.3.4.....2021}\)
\(A=\frac{1}{2021}\)
\(B=\frac{2020.2021-2020.2020}{2020.2019+2020.2}\)
\(B=\frac{2020.\left(2021-2020\right)}{2020.\left(2019+2\right)}\)
\(B=\frac{1}{2021}\)
Từ đó ta thấy 2 biểu thức bằng nhau
![](https://rs.olm.vn/images/avt/0.png?1311)
trả lời:
Đương nhiên là A sẽ lớn hơn B vì các thừa số của A lớn hơn các thừa số của B.
Vậy A>B
học tốt nha bạn!
vì 2020>2019>2009>2000nen a>b
chúc học tốt...................
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!