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giup mình với

Ta có:

\(A=\frac{2021^{2021}+1}{2021^{2022}+1}\Leftrightarrow10A=\frac{2021^{2022}+10}{2021^{2022}+1}=1+\frac{9}{2021^{2022}+1}\)

\(B=\frac{2021^{2022}-1}{2021^{2023}-1}\Leftrightarrow10B=\frac{2021^{2023}-10}{2021^{2023}-1}=1-\frac{9}{2021^{2023}-1}\)

Hay ta đang so sánh: \(\frac{9}{2021^{2022}};\frac{9}{2021^{2023}}\)

Mà \(\frac{9}{2021^{2022}}>\frac{9}{2021^{2023}}\)nên \(\frac{2021^{2021}+1}{2021^{2022}+1}>\frac{2021^{2022}-1}{2021^{2023}-1}\)hay\(A>B\)

Vậy \(A>B\)

Cảm ơn bạn Nguyễn Đăng Nhân nha !!!

Không cần tính, ta thấy : 2022/2021 > 2021/2022
Vậy : 2022/2021*2023 > 2021/2022*2022

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

1-2+3-4+...+2021-2022+2023

=(1-2)+(3-4)+...+(2021-2022)+2023

=(-1)+(-1)+(-1)+...+(-1)+2023

=(-1011)+2023

=1012

\(1-2+3-4+...+2021-2022+2023\)

\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2023\)

\(=-1011+2023=1012\)

\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)

\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)

mà 10^2023+1>10^2022+1

nên A<B

=)

cc

(125.7^2-5.35^2): 2021^2022

= ( 125.49 - 5. 875 ) : 4086462

= ( 6125 - 4375 ) : 4086462

= 4086462 : 1750

= 2335,12114

cảm ơn bạn

thì bạn đi