B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022

B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\) 

B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\) 

B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))

Vậy B > C 

Cứu với ;-;

a) Ta có:

Suy ra: 

Do đó .

Lại có .

Vậy A < B.

Vì từ  có  số

\(2A=\dfrac{2^{2021}-1-1}{2^{2021}-1}=1-\dfrac{1}{2^{2021}-1}\)

\(2B=\dfrac{2^{2022}-1-1}{2^{2022}-1}=1-\dfrac{1}{2^{2022}-1}\)

mà \(2^{2021}-1< 2^{2022}-1\)

nên A<B

$A=\frac{{}^{}-1}{{}^{}-1}$

$\Rightarrow 2A=\frac{2.({}^{}-1)}{{}^{}-1}$

$\Rightarrow 2A=\frac{{}^{}-2}{{}^{}-1}$

$\Rightarrow 2A=\frac{{}^{}-1-1}{{}^{}-1}$

$\Rightarrow 2A=1-\frac{1}{{}^{}-1}$

$B=\frac{{}^{}-1}{{}^{}-1}$

$\Rightarrow 2B=\frac{2.({}^{}-1)}{{}^{}-1}$

Đây dang lớp mấy vây

\(2.A=\frac{2^{2021}-2}{2^{2021}-1}=1-\frac{1}{2^{2021}-1}\)

\(2B=\frac{2^{2022}-2}{2^{2022}-1}=1-\frac{1}{2^{2022}-1}\)

dó \(\frac{1}{2^{2022}-1}< \frac{1}{2^{2021}-1}\Rightarrow1-\frac{1}{2^{2022}-1}>1-\frac{1}{2^{2021}-1}\Rightarrow A< B\)

HT

B/A

\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)

\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)