a) Ta có:

Suy ra: 

Do đó .

Lại có .

Vậy A < B.

B/A

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

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

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

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

mà \(2^{2021}-1< 2^{2022}-1\)

nên A<B

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

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

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

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

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

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

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

\(B=\left(\dfrac{2020}{2}+1\right)+\left(\dfrac{2019}{3}+1\right)+...+\left(\dfrac{1}{2021}+1\right)+1\)

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

=2022(1/2+1/3+...+1/2021+1/2022)

=>B/A=2022

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 

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

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

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

Đặt $B=2023-2022+2021-2020+...+3-2+1$

$B=(2023-2022)+(2021-2020)+...+(3-2)+1$

Đặt $A=(2023-2022)+(2021-2020)+...+(3-2)$

Biểu thức $A$ có số số hạng là: 

$(2023-2):1+1=2022$ (số hạng)

Số nhóm được lập là: 

$2022:2=1011$ (nhóm)

$A=1+1+...+1$    [$1011$ số hạng]

$A=1\times 1011=1011$

$\Rightarrow B=1011+1=1012$

Vậy $B=1012$

ta có 1 công 1 công 1 bằng  dáp án là

1011

=(2023-2022)+(2021-2020)+...+(4-3)=(2-1)

=1+1+1+...+1+1

=1011