\(\left(2^{2021}+2^{2022}\right):2^{2020}=2^{2021}:2^{2020}+2^{2022}:2^{2020}\)

\(=2^{2021-2020}+2^{2022-2020}=2^1+2^2=2+4=6\)

   (2²⁰²¹ + 2²⁰²²) : 2²⁰²⁰ 

= (2 + 2²).2²⁰²⁰ : 2²⁰²⁰

= 2 + 4

= 6

2020/2021<1

2021/2022<1

2022/2023<1

2023/2020=1+1/2020+1/2020+1/2020>1+1/2021+1/2022+1/2023

=>B>2020/2021+2021/2022+2022/2023+1/2021+1/2022+1/2023+1=4

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 

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

a: =2+6*(-1)^2019+2026

=2028-6

=2022

b: \(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}...\cdot\dfrac{625}{624}\)

\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{4^2}{\left(4-1\right)\left(4+1\right)}...\cdot\dfrac{625}{\left(25-1\right)\left(25+1\right)}\)

\(=\dfrac{2\cdot3\cdot4\cdot...\cdot49}{1\cdot2\cdot3\cdot...\cdot48}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot49}{3\cdot4\cdot5\cdot...\cdot50}\)

\(=\dfrac{49}{1}\cdot\dfrac{2}{50}=\dfrac{98}{50}=\dfrac{49}{25}\)

\(=5^{2022}:5^{2021}+5^{2021}:5^{2021}=5+1=6\)

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))

vậy x= 2023