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\(A=\dfrac{2023^{2022+2}}{2023^{2022-1}}=2023^{2024-2021}=2023^3\\ B=\dfrac{2023^{2022}}{2023^{2022-3}}=2023^3\\ \Rightarrow A=B\left(=2023^3\right)\)

A=1x2+2x3+3x4+...+49x50

3A= 3(1.2+2.3+3.4+...+49.50)

3A= 1.2.3+2.3.3+3.4.3+...+49.50.3

3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+49.50.(51-48)

3A= 0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51

3A= 49.50.51

A= 49.50.51/3=41650

B=1x3+3x5+5x7+...+99x101

B=1/1.3 +1/3.5 +...+1/99.101

2B=2/1.3 + 2/3.5 +...+2/99.101

2B=1-1/3+1/3-1/5+...+1/99-1/101

2B=1-1/101

2B=100/101

B=100/101:2=100/202

\(3B=1.3^2+2.3^3+3.3^4+...+2022.3^{2023}+2023.3^{2024}\)

\(2B=3B-B=-3-3^2-3^3-...-3^{2023}+2023.3^{2024}\)

\(2B=2023.3^{2024}-\left(3+3^2+3^3+...+3^{2023}\right)\)

Đặt 

\(C=3+3^2+3^3+...+3^{2023}\)

\(3C=3^2+3^3+3^4+...+3^{2024}\)

\(2C=3C-C=3^{2024}-3\Rightarrow C=\dfrac{3^{2024}-3}{2}\)

\(\Rightarrow2B=2023.3^{2024}-\dfrac{3^{2024}-3}{2}=\)

\(=\dfrac{2.2023.3^{2024}-3^{2024}+3}{2}=\dfrac{4045.3^{2024}+3}{2}\)

\(\Rightarrow B=\dfrac{4045.3^{2024}+3}{4}\)

Bạn nên viết lại đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.

chịu

24^5 K+1

@Bé Bin bạn có biết cách giải không zậy

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022

B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\) 

B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\) 

B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))

Vậy B > C