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NV
4 tháng 1

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

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

4 tháng 1

   (22021 + 22022) : 22020 

= (2 + 22).22020 : 22020

= 2 + 4

= 6

11 tháng 8 2023

\(=2023-1^{2020}+1=2023\)

3 tháng 5 2023

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 

 

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

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

mà \(2^{2021}-1< 2^{2022}-1\)

nên A<B

27 tháng 2 2022

A=22020-122021-1

⇒2A=2.(22020-1)22021-1

⇒2A=22021-222021-1

⇒2A=22021-1-122021-1

⇒2A=1-122021-1

B=22021-122022-1

⇒2B=2.(22021-1)22022-1

24 tháng 7 2021

dễ ợt

 

B/A

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

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

DD
2 tháng 3 2021

a) \(M=2020+2020^2+...+2020^{10}\)

\(M=\left(2020+2020^2\right)+\left(2020^3+2020^4\right)+...+\left(2020^9+2020^{10}\right)\)

\(M=2020\left(1+2020\right)+2020^3\left(1+2020\right)+...+2020^9\left(1+2020\right)\)

\(M=2021\left(2020+2020^3+...+2020^9\right)⋮2021\).

b) Bạn làm tương tự câu a). 

2 tháng 3 2021

b, \(A=2021+2021^2+...+2021^{2020}\)

\(=2021\left(1+2021\right)+...+2021^{2019}\left(1+2021\right)\)

\(=2022\left(2021+...+2021^{2019}\right)⋮2022\)

Vậy ta có đpcm 

Đặt D=2^2021+2^2020+...+2+1

=>2D=2^2022+2^2021+...+2^2+2

=>D=2^2022-1

=>C=2^2022-2^2022+1=1

27 tháng 1 2023

A=20222−20212+20202−20192+...+22−1�=20222-20212+20202-20192+...+22-1

=(2022−2021)(2022+2021)+(2020−2019)(2020+2019)+....+(2−1)(2+1)=(2022-2021)(2022+2021)+(2020-2019)(2020+2019)+....+(2-1)(2+1)

=1.4043+1.4039+.....+1.3=1.4043+1.4039+.....+1.3

Vì từ 3→40433→4043 có 10111011 số

⇒A=(4043+3).10112⇒�=(4043+3).10112

=4046.10112=2023.1011=4046.10112=2023.1011

=2045253