a) \(2023^{2024}\) và \(2023^{2023}\)

vì 2024 > 2023 nên 2023²⁰²⁴ > 2023²⁰²³

Vậy 2023²⁰²⁴ > 2023²⁰²³

b) \(17^{2024}\) và \(18^{2024}\)

vì 17 < 18 nên 17²⁰²⁴ < 18 ²⁰²⁴

Vậy 17²⁰²⁴ < 18²⁰²⁴

a)>

b)<

giúp em với

\(A=\dfrac{2024^{2023}+1}{2024^{2024}+1}\)

\(2024A=\dfrac{2024^{2024}+2024}{2024^{2024}+1}=\dfrac{\left(2024^{2024}+1\right)+2023}{2024^{2024}+1}=\dfrac{2024^{2024}+1}{2024^{2024}+1}+\dfrac{2023}{2024^{2024}+1}=1+\dfrac{2023}{2024^{2024}+1}\)

\(B=\dfrac{2024^{2022}+1}{2024^{2023}+1}\)

\(2024B=\dfrac{2024^{2023}+2024}{2024^{2023}+1}=\dfrac{\left(2024^{2023}+1\right)+2023}{2024^{2023}+1}=\dfrac{2024^{2023}+1}{2024^{2023}+1}+\dfrac{2023}{2024^{2023}+1}=1+\dfrac{2023}{2024^{2023}+1}\)

Vì \(2024>2023=>2024^{2024}>2024^{2023}\)

\(=>2024^{2024}+1>2024^{2023}+1\)

\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)

\(=>A< B\)

\(#PaooNqoccc\)

dễ

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

Ta có : \(N=2022.2024\)

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

\(N=2023^2+2023-2023-1\)

\(N=2023^2-1\)

Mà : \(M=2023.2023=2023^2\)

\(\Rightarrow M>N\)

\(A=\dfrac{10^{2024}+1}{10^{2023}+1}=\dfrac{10\left(10^{2023}+1\right)}{10^{2023}+1}-\dfrac{9}{10^{2023}+1}=1-\dfrac{9}{10^{2023}+1}\)

\(B=\dfrac{10^{2023}+1}{10^{2022}+1}=\dfrac{10\left(10^{2022}+1\right)}{10^{2022}+1}-\dfrac{9}{10^{2022}+1}=1-\dfrac{9}{10^{2022}+1}\)

Vì \(\dfrac{9}{10^{2023}+1}< \dfrac{9}{10^{2022}+1}\)

\(\Rightarrow A>B\)

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