K
Khách

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

15 tháng 6 2019

         \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{2019}-1\right)\)

\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-2018}{2019}\)

\(=\frac{-1.1}{2}.\frac{-1.2}{3}.\frac{-1.3}{4}....\frac{-1.2018}{2019}\)

\(=\left(-1\right)^{2018}.\frac{1.2.3....2018}{2.3.4....2019}\)

\(=1.\frac{1}{2019}=\frac{1}{2019}\)

17 tháng 6 2019

xin loi ban nha , minh bam nham dung roi day

3 tháng 8 2023

Ok em, để olm.vn giúp em nhá: 

A = \(\dfrac{1}{2}\):3 + \(\dfrac{1}{3}\):4 + \(\dfrac{1}{4}\):5+...+\(\dfrac{1}{2018}\):2019 + \(\dfrac{1}{2019}\): 2020

A=\(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+..+\dfrac{1}{2018}\times\dfrac{1}{2019}+\dfrac{1}{2019}\times\dfrac{1}{2020}\)

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+....+ \(\dfrac{1}{2018}\) - \(\dfrac{1}{2019}\)\(\dfrac{1}{2019}\) - \(\dfrac{1}{2020}\)

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{2020}\)

A = \(\dfrac{1009}{2020}\)

3 tháng 8 2023

Giúp mình nhé 

 

11 tháng 2 2019

A = (-1)(-1)^2(-1)^3...(-1)^2019

A = (-1)^1+2+3+...+2019

A = (-1)^2039190

A = 1

S = 1.2.3 + 2.3.4 + 3.4.5 + ... + 2018.2019.2020

4S = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + .... + 2018.2019.2020.4

4S = 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + ... + 2018.2019.2020.(2021 - 2017)

4S = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 2018.2019.2020.2021 - 2017.2018.2019

4S = 2018.2019.2020.2021

S = 2018.2019.2020.2021 : 4 = ...

cảm ơn bạn nhiều nhé

B/A

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

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

3 tháng 4 2019

\(B=\frac{2019}{1}+\frac{2018}{2}+\frac{2017}{3}+......+\frac{1}{2019}\)

\(=\left(\frac{2018}{2}+1\right)+\left(\frac{2017}{3}+1\right)+.....+\left(\frac{1}{2019}+1\right)+1\)

\(=\frac{2020}{2}+\frac{2020}{3}+\frac{2020}{4}+.....+\frac{2020}{2019}+\frac{2020}{2020}\)

\(=2020\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2020}\right)\)

\(=2020A\)

\(\Rightarrow\frac{A}{B}=\frac{A}{2020A}=\frac{1}{2020}\)