17/21>17/29>15/29 (tự KL) b)(chép đề) B= (20+21)/(21+22)=41/43<1 (chép đề sai). Xét A= 20/21+21/22=1-1/21+1-1/22=1+1-(1/21+1/22). Ta thấy (1/21+1/22)<1 nên 1-(1/21+1/22)>0

17/21>17/29>15/29 (tự KL)
b)(chép đề)
B= (20+21)/(21+22)=41/43<1 (chép đề sai).
Xét A= 20/21+21/22=1-1/21+1-1/22=1+1-(1/21+1/22). 
Ta thấy (1/21+1/22)<1 
nên 1-(1/21+1/22)>0
Vậy 1+1-(1/21+1/22)>1+0>1
Vậy A>B

Lời giải:
$10A=\frac{10^{2021}-10}{10^{2021}-1}=\frac{10^{2021}-1-9}{10^{2021}-1}$

$=1-\frac{9}{10^{2021}-1}>1$

$10B=\frac{10^{2022}+10}{10^{2022}+1}=\frac{10^{2022}+1+9}{10^{2022}+1}$

$=1+\frac{9}{10^{2022}+1}<1$

$\Rightarrow 10A> 1> 10B$

Suy ra $A> B$

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: \(B=2020.2021.2022=\left(2021-1\right).\left(2021+1\right).2021=\left(2021-1\right)^2.2021< 2021^2.2021=A\)

ghi rõ đề bài ra nhanh lên

`# \text {DNamNgV}`

\(A=1+2+2^2+...+2^{2021}\text{ và }B=2^{2022}\)

Ta có:

\(A=1+2+2^2+...+2^{2021}\\ \Rightarrow2A=2+2^2+2^3+...+2^{2022}\\\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2022}\right)-\left(1+2+2^2+...+2^{2021}\right)\\ \Rightarrow A=2+2^2+2^3+...+2^{2022}-1-2-2^2-...-2^{2021}\\ \Rightarrow A=2^{2022}-1\)

Vì \(2^{2022}-1< 2^{2022}\)

\(\Rightarrow A< B.\)

A=B

Ta có:

\(10A=\dfrac{10\left(10^{2020}+1\right)}{10^{2021}+1}=\dfrac{10^{2021}+10}{10^{2021}+1}=1+\dfrac{9}{10^{2021}+1}\)

\(10B=\dfrac{10\left(10^{2021}+1\right)}{10^{2022}+1}=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)

⇒ \(10A>10B\) ( vì \(\dfrac{9}{10^{2021}+1}>\dfrac{9}{10^{2022}+1}\) )

Suy ra:  \(A>B\)