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AH
Akai Haruma
Giáo viên
15 tháng 4 2023

Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+....+\frac{1}{\frac{2023.2024}{2}}$

$=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2023.2024}$

$=2(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2024-2023}{2023.2024})$

$=2(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2023}-\frac{1}{2024})$

$=2(\frac{1}{3}-\frac{1}{2024})=\frac{2021}{3036}$

4 tháng 3

=23.4+24.5+...+22023.2024

=2(4−33.4+5−44.5+...+2024−20232023.2024)

=2(13−14+14−15+....+12023−12024)

=2(13−12024)=20213036
 

AH
Akai Haruma
Giáo viên
31 tháng 1

Lời giải:

\(C=(\frac{1}{2^2}-1)(\frac{1}{3^2}-1)(\frac{1}{4^2}-1)....(\frac{1}{2023^2}-1)\)

\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}....\frac{1-2023^2}{2023^2}\)

\(=\frac{(2^2-1)(3^2-1)(4^2-1)....(2023^2-1)}{2^2.3^2.4^2....2023^2}\)

\(=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)....(2023-1)(2023+1)}{2^2.3^2.4^2....2023^2}\)

\(=\frac{1.3.2.4.3.5.....2022.2024}{(2.3.4...2023)(2.3.4...2023)}\)

\(=\frac{(1.2.3...2022)(3.4.5....2024)}{(2.3...2023)(2.3.4...2023)}\)

\(=\frac{1}{2023}.\frac{2024}{2}=\frac{1012}{2023}\)

 

 

31 tháng 1

\(\dfrac{1012}{2023}\)

TA
28 tháng 10 2023

4072299/4048

1 tháng 11 2023

cho mik câu trả lời cụ thể đc k bn

\(B=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^4+...-\dfrac{1}{2022}+\dfrac{1}{2023}\\ \Rightarrow B=\dfrac{2}{2^2}-\dfrac{1}{2^2}+\dfrac{2}{2^4}-\dfrac{1}{2^4}+...+\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)

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

\(\Rightarrow B=\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+...+\dfrac{1}{2^{2024}}\\ \Rightarrow2^2B=\dfrac{2^{2024}}{2^{2024}}+\dfrac{2^{2022}}{2^{2024}}+...+\dfrac{2^2}{2^{2024}}\)

\(\Rightarrow4B-B=\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\\ \Rightarrow3B=1-\left(\dfrac{2}{2^{2024}}+\dfrac{1}{2^{2024}}\right)\)

\(\Rightarrow3B=1-\dfrac{3}{2^{2024}}\\ \Rightarrow B=\dfrac{1-\dfrac{3}{2^{2024}}}{3}\)

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

 

a: =-3/4-1/4+2/7+5/7+2023/2024

=-1+1+2023/2024=2023/2024

b: 2/3x=2/7

=>x=2/7:2/3=3/7

c; =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5

=>x=3/5:2/3=3/5*3/2=9/10