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11 tháng 7 2017

\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)

\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)

\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)

\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)

\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)

\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)

\(B=\dfrac{4.9.16.100}{3.8.15.99}\)

\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)

\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)

\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)

25 tháng 5 2022

\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

25 tháng 5 2022

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17 tháng 10 2023

\(A=\dfrac{1}{2}\left(\dfrac{2.2}{1.3}\right).\left(\dfrac{3.3}{2.4}\right)...\left(\dfrac{2020.2020}{2019.2021}\right)\)

\(=\dfrac{1.2.2.3.3...2020.2020}{1.2.2.3.3.4.4...2019.2021}\)

\(=\dfrac{1}{2021}\)

17 tháng 10 2023

\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{2019\cdot2021}\right)\)

\(A=\dfrac{1}{2}\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\left(1+\dfrac{1}{4^2-1}\right)...\left(1+\dfrac{1}{2020^2-1}\right)\)

\(A=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\cdot\left(3+1\right)}...\left(\dfrac{2020^2}{\left(2020-1\right)\cdot\left(2020+1\right)}\right)\)

\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2020}{2021}\)

\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}\)

\(A=\dfrac{1}{2}\cdot2020\cdot\dfrac{2}{2021}=\dfrac{2020}{2021}\)

\(=\dfrac{1}{2}\cdot\dfrac{2^2-1+1}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2-1+1}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2016^2-1+1}{\left(2016-1\right)\left(2016+1\right)}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2016}{2015}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2016}{2017}\)

\(=\dfrac{1}{2}\cdot2016\cdot\dfrac{2}{2017}=\dfrac{2016}{2017}\)

30 tháng 4 2018

https://hoc24.vn/hoi-dap/question/598367.html

19 tháng 9 2023

P = (1 + \(\dfrac{1}{1.3}\)).(1 + \(\dfrac{1}{2.4}\)).(1 + \(\dfrac{1}{3.5}\))....(1 + \(\dfrac{1}{2020.2022}\))

P = \(\dfrac{1.3+1}{1.3}\)\(\dfrac{2.4+1}{2.4}\).\(\dfrac{3.5+1}{3.5}\)....\(\dfrac{2020.2022+1}{2020.2022}\)

P=\(\dfrac{\left(2-1\right)\left(2+1\right)+1}{1.3}\).\(\dfrac{\left(3-1\right)\left(3+1\right)+1}{2.4}\)...\(\dfrac{\left(2021+1\right).\left(2022-1\right)+1}{2020.2022}\)

P = \(\dfrac{2.2}{1.3}\).\(\dfrac{3.3}{2.4}\).\(\dfrac{4.4}{3.5}\)....\(\dfrac{2021.2021}{2020.2022}\)

P = \(\dfrac{2.2021}{2022}\)

P = \(\dfrac{2021}{1011}\) 

 

3 tháng 4 2017

Ta có:
\(A=\dfrac{1}{2}.\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{2016^2}{2015.2017}\)
\(A=\dfrac{1}{2}.\dfrac{2^2}{3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{2016^2}{2015.2017}\)
\(A=\left(\dfrac{2.3.4...2016}{2.3.4.5...2015}\right).\left(\dfrac{2.3.4...2016}{2.3.4.5...2017}\right)\)
\(A=2016.\dfrac{1}{2017}=\dfrac{2016}{2017}\)

4 tháng 4 2017

Có chắc là bạn làm đúng không vậy???

Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)

\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)

Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)

24 tháng 4 2021

bn lm sai rồi

AH
Akai Haruma
Giáo viên
2 tháng 1 2019

Lời giải:

Xét tổng quát:

\(1+\frac{1}{k(k+2)}=\frac{k(k+2)+1}{k(k+2)}=\frac{(k+1)^2}{k(k+2)}\)

Thay $k=1,2,....,2015$ ta có:

\(1+\frac{1}{1.3}=\frac{2^2}{1.3}\)

\(1+\frac{1}{2.4}=\frac{3^2}{2.4}\)

\(1+\frac{1}{3.5}=\frac{4^2}{3.5}\)

\(1+\frac{1}{4.6}=\frac{5^2}{4.6}\)

.............

\(1+\frac{1}{2015.2017}=\frac{2016^2}{2015.2017}\)

Nhân theo vế:

\(\Rightarrow A=\frac{1}{2}\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2015.2017}\right)\)

\(=\frac{1}{2}.\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}....\frac{2016^2}{2015.2017}\)

\(=\frac{(1.2.3...2016)^2}{(1.2.3...2015)(2.3.4...2017)}=\frac{(1.2.3...2016)(2.3....2016)}{(1.2.3...2015)(2.3.4...2017)}=2016.\frac{1}{2017}=\frac{2016}{2017}\)