K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

\(=\dfrac{1-2^2}{2^2}\cdot\dfrac{1-3^2}{3^2}\cdot...\cdot\dfrac{1-2023^2}{2023^2}\)

\(=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot...\cdot\dfrac{2023^2-1}{2023^2}\)

\(=\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{2}{3}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2022}{2023}\cdot\dfrac{2024}{2023}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2022}{2023}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2024}{2023}\)

\(=\dfrac{1}{2023}\cdot\dfrac{2024}{2}=\dfrac{1012}{2023}\)

5 tháng 4

=1012/2023

 

1:

a: =23/27-11/17+4/27+28/17

=23/27+4/27+28/17-11/17

=1+1=2

b: \(=\dfrac{2}{3}\cdot\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{2}{9}\)

=2/3-2/9

=6/9-2/9

=4/9

c: \(=\dfrac{11}{5}\cdot\dfrac{7}{3}-\dfrac{1}{3}\cdot\dfrac{11}{5}\)

=11/5(7/3-1/3)

=11/5*2

=22/5

d: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2024}{2023}=\dfrac{2024}{2}=1012\)

e: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2022}{2023}=\dfrac{1}{2023}\)

 

18 tháng 8 2023

thiếu rồi bạn ơi

3 tháng 5 2023

\(A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{2023}-\left(\dfrac{1}{2}\right)^{2024}\)

\(A=\dfrac{2}{2^2}-\dfrac{1}{2^2}+\dfrac{2}{2^4}-\dfrac{1}{2^4}+...+\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)

\(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{2024}}\)

\(A=\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+\dfrac{2^{2018}}{2^{2024}}+...+\dfrac{1}{2^{2024}}\)

\(2^2A=\dfrac{2^{2024}}{2^{2024}}+\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+...+\dfrac{2^2}{2^{2024}}\)

\(\Rightarrow4A-A=3A=1-\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)

\(3A=1-\left(\dfrac{2}{2^{2024}}+\dfrac{1}{2^{2024}}\right)\)

\(3A=1-\dfrac{3}{2^{2024}}\)

\(A=\dfrac{1-\dfrac{3}{2^{2024}}}{3}\)

\(A=\dfrac{3\left(\dfrac{1}{3}-\dfrac{1}{2^{2024}}\right)}{3}\)

\(A=\dfrac{1}{3}-\dfrac{1}{2^{2024}}\)

3 tháng 5 2023

giúp mk vs các bn. chiều nay mk phải nộp r

19 tháng 7 2023

42 : x + 36 : x = 6

19 tháng 7 2023

TH1

42:x=6

x= 42 :6 

X= 7

TH 2

36:x = 6

X = 36: 6

X= 6

24 tháng 9

Ta có: C = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/2021.2022.2023

=> C = 1/2. (3-1/1.2.3 + 4-2/2.3.4 + 5-3/3.4.5 + ... + 2023-2021/2021.2022.2023

=> C = 1/2. (1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/2021.2022 - 1/2022.2023)

=> C = 1/2. (1/1.2 - 1/2022.2023)

- Phần còn lại bạn tự tính chứ số to quá

24 tháng 7 2021

24^5 K+1

24 tháng 7 2021

@Bé Bin bạn có biết cách giải không zậy

 

22 tháng 3 2023

Có : `1/2^2<1/(1*2)`

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\\ ...\\ \dfrac{1}{2023^2}< \dfrac{1}{2022\cdot2023} \)

nên \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2023^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\\ =1-\dfrac{1}{2023}=\dfrac{2022}{2023}< 1\\ \Rightarrow B< 1\left(đpcm\right)\)

16 tháng 4 2023

=> 4S = 1 + 2/4 + 3/4^2 +...+ 2023/4^2022

=> 4S-S = 1 + (2/4-1/4) + (3/4^2 - 2/4^2) +...+ (2023/4^2022 - 2022/4^2022) - 2023/4^2023

=> 3S = 1 + 1/4 + 1/4^2 +...+ 1/4^2022 - 2023/4^2023

=> 12S = 4 + 1 + 1/4 +... + 1/4^2021 - 2023/4^2022

=> 12S - 3S = 4 + (1-1) + (1/4-1/4) +... + (1/4^2021 - 1/4^2021)  - 1/4^2022 - 2023/4^2022 + 2023/4^2023

=> 9S = 4 -  1/4^2022 - 2023/4^2022 + 2023/4^2023

= 4- 2024/4^2022 + 2023/4^2023

Do 2024/4^2022 > 2024/4^2023 > 2023/4^2023 nên - 2024/4^2022 + 2023/4^2023 < 0

=> 9S < 4 < 9/2

=> S < 1/2 (đpcm)

30 tháng 8 2023

Cho S=1+3+3^2+....+3^2023

Chứng tỏ S chia hết cho 4

16 tháng 4 2023

Ta có S = \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2023}{4^{2023}}\)

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

4S - S = ( \(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2023}{4^{2022}}\) ) - ( \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2023}{4^{2023}}\))

3S = 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}-\dfrac{2023}{4^{2023}}\)

Đặt A = 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}\)

4A = 4 + 1 + \(\dfrac{1}{4}+...+\dfrac{1}{4^{2021}}\)

4A - A = ( 4 + 1 + \(\dfrac{1}{4}+...+\dfrac{1}{4^{2021}}\)) - ( 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}\))

3A = 4 - \(\dfrac{1}{4^{2022}}\)

A = ( 4 - \(\dfrac{1}{4^{2022}}\)) : 3 = \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\)

⇒ 3S = \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\) - \(\dfrac{2023}{4^{2023}}\)

S = ( \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\) - \(\dfrac{2023}{4^{2023}}\)) : 3 = \(\dfrac{4}{9}-\dfrac{1}{4^{2022}\cdot3^2}-\dfrac{1}{4^{2023}\cdot3}< \dfrac{4}{9}< \dfrac{1}{2}\)

Vậy S < \(\dfrac{1}{2}\)

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