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

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

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

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

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

B = \(1-\dfrac{1}{2025}\)   \(A=1-\dfrac{1}{2024}\)

Vì \(\dfrac{1}{2025}< \dfrac{1}{2024}\)

Nên B>A

Ta có :

\(\dfrac{2023}{2024}\)=\(\dfrac{2024-1}{2024}\)=\(\dfrac{2024}{2024}\)-\(\dfrac{1}{2024}\)=1-\(\dfrac{1}{2024}\)

\(\dfrac{2024}{2025}\)=\(\dfrac{2025-1}{2025}\)=\(\dfrac{2025}{2025}\)-\(\dfrac{1}{2025}\)=1=\(\dfrac{1}{2025}\)

Ta thấy: \(\dfrac{1}{2024}\) lớn hơn \(\dfrac{1}{2025}\)

Nên : \(\dfrac{2023}{2024}\) lớn hơn \(\dfrac{2024}{2025}\)

⇒A lớn hơn B

A>B

bạn có thể giải chi tiết được không ạ?

Câu b thì gg search nhé

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)<

\(8A=\dfrac{8^{2022}+16}{8^{2022}+2}=1+\dfrac{14}{8^{2022}+2}\)

\(8B=\dfrac{8^{2024}+16}{8^{2024}+2}=1+\dfrac{14}{8^{2024}+2}\)

Vì \(\dfrac{14}{8^{2022}+2}>\dfrac{14}{8^{2024}+2}\)

=> 8A>8B

=> A>B

thanks

b) \(M=\dfrac{10^{2023}+1}{10^{2024}+1}< 1\) ( Vì tử < mẫu )

Ta có: \(M=\dfrac{10^{2023}+1}{10^{2024}+1}< \dfrac{10^{2023}+1+9}{10^{2024}+1+9}=\dfrac{10^{2023}+10}{10^{2024}+10}=\dfrac{10.\left(10^{2022}+1\right)}{10.\left(10^{2023}+1\right)}=\dfrac{10^{2022}+1}{10^{2023}+1}=N\)

Vì \(\dfrac{10^{2023}+1}{10^{2024}+1}< \dfrac{10^{2022}+1}{10^{2023}+1}\) nên \(M< N\)

\(C=\dfrac{2^{2024}-3}{2^{2023}-1}=\dfrac{2.2^{2023}-2-1}{2^{2023}-1}=\dfrac{2\left(2^{2023}-1\right)-1}{2^{2023}-1}=2-\dfrac{1}{2^{2023}-1}\)

\(D=\dfrac{2^{2023}-3}{2^{2022}-1}=\dfrac{2.2^{2022}-2-1}{2^{2022}-1}=\dfrac{2\left(2^{2022}-1\right)-1}{2^{2022}-1}=2-\dfrac{1}{2^{2022}-1}\)

Ta có

\(2^{2023}>2^{2022}\Rightarrow2^{2023}-1>2^{2022}-1\)

\(\Rightarrow\dfrac{1}{2^{2023}-1}< \dfrac{1}{2^{2022}-1}\Rightarrow2-\dfrac{1}{2^{2023}-1}>2-\dfrac{1}{2^{2022}-1}\)

\(\Rightarrow C>D\)