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 

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

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

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

Đây dang lớp mấy vây

Cứu với ;-;

help me !!!!!!!

\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)

= 1.495676488x10^-3

a) Ta có:

Suy ra: 

Do đó .

Lại có .

Vậy A < B.