2020/2021<1

2021/2022<1

2022/2023<1

2023/2020=1+1/2020+1/2020+1/2020>1+1/2021+1/2022+1/2023

=>B>2020/2021+2021/2022+2022/2023+1/2021+1/2022+1/2023+1=4

...

\(\dfrac{C_n^k}{\left(k+1\right)\left(k+2\right)}=\dfrac{n!}{\left(k+1\right)\left(k+2\right).k!\left(n-k\right)!}=\dfrac{1}{\left(n+1\right)\left(n+2\right)}.\dfrac{\left(n+2\right)!}{\left(n+2-\left(k+2\right)\right)!\left(k+2\right)!}\)

\(=\dfrac{1}{\left(n+1\right)\left(n+2\right)}.C_{n+2}^{k+2}\)

Đặt tổng trên là A

\(\Rightarrow A=\dfrac{-1.C_{2024}^3}{2023.2024}+\dfrac{2.C_{2024}^4}{2023.2024}+\dfrac{-3.C_{2024}^5}{2023.2024}+...+\dfrac{2022.C_{2024}^{2024}}{2023.2024}\)

\(=\dfrac{1}{2023.2024}\left(-1.C_{2024}^3+2.C_{2024}^4+...+2022.C_{2024}^{2024}\right)=\dfrac{1}{2023.2024}.B\)

Xét \(C=-2.\left(-C_{2024}^3+C_{2024}^4-C_{2024}^5+...+C_{2024}^{2024}\right)\)

\(\Rightarrow B-C=-3C_{2024}^3+4C_{2024}^4-5C_{2024}^5+...+2024.C_{2024}^{2024}\)

Ta có:

\(k.C_n^k=\dfrac{n!.k}{\left(n-k\right)!.k!}=n.\dfrac{\left(n-1\right)!}{\left(\left(n-1\right)-\left(k-1\right)\right)!.\left(k-1\right)!}=n.C_{n-1}^{k-1}\)

\(\Rightarrow B-C=-2024.C_{2023}^2+2024C_{2023}^3+...+2024.C_{2023}^{2023}\)

\(=-2024\left(C_{2023}^2-C_{2023}^3+...-C_{2023}^{2023}\right)\)

Xét khai triển:

\(\left(1-x\right)^k=C_k^0-xC_k^1+x^2C_k^2+...+\left(-1\right)^kx^k.C_k^k\)

Thay \(k=2024\); \(x=1\)

\(\Rightarrow0=C_{2024}^0-C_{2024}^1+C_{2024}^2-C_{2024}^3+...+C_{2024}^{2024}\)

\(\Rightarrow-C_{2024}^3+...+C_{2024}^{2024}=C_{2024}^1-C_{2024}^2-1\)

\(\Rightarrow C=-2\left(C_{2024}^1-C_{2024}^2-1\right)=-2\left(2023-C_{2024}^2\right)\)

Thay \(k=2023;x=1\)

\(\Rightarrow0=C_{2023}^0-C_{2023}^1+C_{2023}^2+...-C_{2023}^{2023}\)

\(\Rightarrow C_{2023}^2-C_{2023}^3+...-C_{2023}^{2023}=C_{2023}^1-1=2022\)

\(\Rightarrow B-C=-2024.2022\)

\(\Rightarrow B=C-2022.2024=-2\left(2023-C_{2024}^2\right)-2022.2024\)

\(=-2.2023+2023.2024-2022.2024\)

\(=-2022\)

\(\Rightarrow A=\dfrac{-2022}{2023.2024}\)

\(\dfrac{2023}{2022}=\dfrac{2022}{2022}+\dfrac{1}{2022}=1+\dfrac{1}{2022}\)

\(\dfrac{2021}{2020}=\dfrac{2020}{2020}+\dfrac{1}{2020}=1+\dfrac{1}{2020}\)

\(\dfrac{1}{2022}< \dfrac{1}{2020}\)

\(\Rightarrow\dfrac{2023}{2022}< \dfrac{2021}{2020}\)

\(\dfrac{2023}{2022}=1+\dfrac{1}{2022}\)

\(\dfrac{2021}{2020}=1+\dfrac{1}{2020}\)

mà \(\dfrac{1}{2022}< \dfrac{1}{2020}\)

nên \(\dfrac{2023}{2022}< \dfrac{2021}{2020}\)

20212020=20202020+12020=1+1202020212020=20202020+12020=1+12020

⇒20232022<20212020

giúp vơi

khó difficult

2022

2022

2022

3x(8^2-2(2^5-1))=2022

=>3x(64-2*31)=2022

=>3x=1011

=>x=337

3x[8² - 2(2⁵ - 1)] = 2022

3x[64 - 2(32 - 1)] = 2022

3x(64 - 2.31) = 2022

3x(64 - 62) = 2022

3x.2 = 2022

6x = 2022

x = 2022 : 6

x = 337

A<B

\(2022A=2022+2022^2+2022^3+2022^4+...+2022^{2018}\)

\(2021A=2022A-A=2022^{2018}-1\Rightarrow A=\dfrac{2022^{2018}-1}{2021}\)

\(\Rightarrow A< B\)