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Sửa đề: A=(1+1/1*3)(1+1/2*4)*...*(1+1/2019*2021)
\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2020^2}{\left(2020-1\right)\left(2020+1\right)}\)
\(=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}=2020\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)
\(A=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(1+\dfrac{1}{99.101}\right)\)
\(A=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}....\dfrac{10000}{99.101}\)
\(A=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}....\dfrac{100^2}{99.101}\)
\(A=\dfrac{2.3.4...100}{1.2.3....99}.\dfrac{2.3.4....100}{3.4.5....101}\)
\(A=100.\dfrac{2}{101}=\dfrac{200}{101}\)
Vậy A = \(\dfrac{200}{101}\)
Chúc học tốt!!
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: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)
\(A=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{24}\right).\left(1+\dfrac{1}{3.5}\right).....\left(1+\dfrac{1}{2014.2016}\right)\)
\(A=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.....\dfrac{4060225}{2014.2016}\)
\(A=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{2015^2}{2014.2016}\)
\(A=\dfrac{2.3.4.5...2015}{1.2.3...2014}.\dfrac{2.3.4...2015}{3.4.5...2016}\)
\(A=2015.\dfrac{2}{2016}=2015.\dfrac{1}{1008}=\dfrac{2015}{1008}\)
Vậy \(A=\dfrac{2015}{1008}\)
(1+1/1.3)...(1+1/2016.2018)
=(1.3+1/1.3)...(2016.2018+1/2016.2018)
=(2.2/1.3)...(2017.2017/2016.2018)
=(2...2017).(2..2017)/(1.2.....2016).(3...2018)
=2017.2/2018
=2017.2/1006.2
=2017/1006
thông cảm nha :))
Ta có \(1+\dfrac{1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{\left(k-1\right)\left(k+1\right)+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2-1+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2}{\left(k-1\right)\left(k+1\right)}\).
Từ đó \(1+\dfrac{1}{1.3}=\dfrac{2^2}{1.3}\); \(1+\dfrac{1}{2.4}=\dfrac{3^2}{2.4}\); \(1+\dfrac{1}{3.5}=\dfrac{4^2}{3.5}\); \(1+\dfrac{1}{4.6}=\dfrac{5^2}{4.6}\);...; \(1+\dfrac{1}{2022.2024}=\dfrac{2023^2}{2022.2024}\).
Suy ra \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2022.2024}\right)\)
\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}...\dfrac{2023^2}{2022.2024}\)
\(=\dfrac{2.2023}{2024}\) \(=\dfrac{2023}{1012}\)