`a, A = 3020 xx 3110 - 5 = 3020 xx 3109 + 3020 - 5`

`= 3020 xx 3109 + 3015 = B`.

`b, B = (2022-2)(2022+2) = 2022^2-4 < 2022^2 = A.`

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

giúp em với

A>B

bạn có thể giải chi tiết được không ạ?

Tham khảo:

Sửa đề: 6-8+10-12+...+2022-2024

Số số hạng của dãy số là: \(\frac{2024-6}{2}+1=\frac{2018}{2}+1=1009+1=1010\) (số)

6-8+10-12+...+2022-2024

=(6-8)+(10-12)+...+(2022-2024)

=(-2)+(-2)+...+(-2)

\(=\left(-2\right)\cdot\frac{1010}{2}=-1010\)

Sửa đề: 6-8+10-12+...+2018-2020+2022-2024

Số số hạng của dãy số là:

\(\frac{2024-6}{2}+1=\frac{2018}{2}+1=\frac{2020}{2}=1010\) (số)

6-8+10-12+...+2018-2020+2022-2024

=(6-8)+(10-12)+...+(2018-2020)+(2022-2024)

=(-2)+(-2)+...+(-2)+(-2)

\(=\left(-2\right)\cdot\frac{1010}{2}=-1010\)

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

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

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

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

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

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

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

\(=>A< B\)

\(#PaooNqoccc\)

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