Tính:
A=\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right)....\left(1+\dfrac{1}{2014.2016}\right)\)
Lưu ý: Dấu . là dấu nhân
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MC
24 tháng 6 2018
\(C=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{3}{3.5}\right)...\left(1+\dfrac{2014}{2016}\right)\)
\(C=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.....\dfrac{4060225}{2014.2016}\)
\(C=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{2015.2015}{2014.2016}\)
\(C=\dfrac{2.2.3.3.4.4.....2015.2015}{1.3.2.4.3.5.....2014.2016}\)
\(C=\dfrac{2.\left(3.2\right)\left(4.3\right).....\left(2015.2014\right).2015}{1.\left(3.2\right)\left(4.3\right).....\left(2015.2014\right).2016}\)\(\)
\(C=\dfrac{2.2015}{1.2016}\)
\(C=\dfrac{4030}{2016}\)\(=1\dfrac{2014}{2016}\).
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 2 2021
Chắc ngoặc đầu tiên là \(\left(1+\dfrac{1}{1.3}\right)\) đúng ko bạn (mặc dù đề như bạn thì vẫn tính được)
\(1+\dfrac{1}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)+1}{n\left(n+2\right)}=\dfrac{n^2+2n+1}{n\left(n+2\right)}=\dfrac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(\Rightarrow C=\dfrac{2^2.3^2...2015^2}{1.3.2.4...2014.2016}=\dfrac{2.3...2015}{1.2...2014}.\dfrac{2.3...2015}{3.4...2016}=\dfrac{2015}{1}.\dfrac{2}{2016}=\dfrac{2015}{1008}\)
9 tháng 3 2021
\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
25 tháng 5 2022
\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
16 tháng 6 2023
Sửa đề: A=(1+1/1*3)(1+1/2*4)*...*(1+1/2019*2021)
\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2020^2}{\left(2020-1\right)\left(2020+1\right)}\)
\(=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}=2020\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)
\(A=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{24}\right).\left(1+\dfrac{1}{3.5}\right).....\left(1+\dfrac{1}{2014.2016}\right)\)
\(A=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.....\dfrac{4060225}{2014.2016}\)
\(A=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{2015^2}{2014.2016}\)
\(A=\dfrac{2.3.4.5...2015}{1.2.3...2014}.\dfrac{2.3.4...2015}{3.4.5...2016}\)
\(A=2015.\dfrac{2}{2016}=2015.\dfrac{1}{1008}=\dfrac{2015}{1008}\)
Vậy \(A=\dfrac{2015}{1008}\)