Tham khảo:

Lời giải:

Ta thấy: $\frac{2021^2+1}{2021}=2021+\frac{1}{2021}< 2022< 2022+\frac{1}{2022}=\frac{2022^2+1}{2022}$

$\Rightarrow \frac{2021}{2021^2+1}> \frac{2022}{2022^2+1}$

Ta có:

\(2023^{2022}=2023\cdot2023^{2021}\)

\(2022^{2022}+2022^{2021}=2022^{2021}\cdot\left(2022+1\right)=2023\cdot2022^{2021}\)

Mà: \(2023>2022\)

\(\Rightarrow2023^{2021}>2022^{2021}\)

\(\Rightarrow2023^{2021}\cdot2023>2022^{2021}\cdot2023\)

\(\Rightarrow2023^{2022}>2022^{2022}+2022^{2021}\) 

Vậy: ... 

a: \(B=\dfrac{154}{155+156}+\dfrac{155}{155+156}\)

\(\dfrac{154}{155}>\dfrac{154}{155+156}\)

\(\dfrac{155}{156}>\dfrac{155}{155+156}\)

=>154/155+155/156>(154+155)/(155+156)

=>A>B

b: \(C=\dfrac{2021+2022+2023}{2022+2023+2024}=\dfrac{2021}{6069}+\dfrac{2022}{6069}+\dfrac{2023}{6069}\)

2021/2022>2021/6069

2022/2023>2022/2069

2023/2024>2023/6069

=>D>C

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 

So sánh 3 số hay sao đây bạn?

\(\dfrac{2021}{2022}=\dfrac{2020}{2021}\)

\(\dfrac{2021}{2022}\) và \(\dfrac{2020}{2021}\)

\(\dfrac{2021}{2022}=1-\dfrac{1}{2022}\)

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

\(\text{Vì }\)\(\dfrac{1}{2022}>\dfrac{1}{2021}=>1-\dfrac{1}{2022}>1-\dfrac{1}{2021}=>\dfrac{2021}{2022}>\dfrac{2020}{2021}\)

2021/2022 > 2020/2021