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Ngọc Nhi
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Mặc Chinh Vũ
24 tháng 6 2018 lúc 16:19

\(C=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{3}{3.5}\right)...\left(1+\dfrac{2014}{2016}\right)\)

\(C=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.....\dfrac{4060225}{2014.2016}\)

\(C=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{2015.2015}{2014.2016}\)

\(C=\dfrac{2.2.3.3.4.4.....2015.2015}{1.3.2.4.3.5.....2014.2016}\)

\(C=\dfrac{2.\left(3.2\right)\left(4.3\right).....\left(2015.2014\right).2015}{1.\left(3.2\right)\left(4.3\right).....\left(2015.2014\right).2016}\)\(\)

\(C=\dfrac{2.2015}{1.2016}\)

\(C=\dfrac{4030}{2016}\)\(=1\dfrac{2014}{2016}\).

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Nguyễn Thị Thanh Hương
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Phúc Trần
7 tháng 3 2018 lúc 19:26

\(A=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{24}\right).\left(1+\dfrac{1}{3.5}\right).....\left(1+\dfrac{1}{2014.2016}\right)\)

\(A=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.....\dfrac{4060225}{2014.2016}\)

\(A=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{2015^2}{2014.2016}\)

\(A=\dfrac{2.3.4.5...2015}{1.2.3...2014}.\dfrac{2.3.4...2015}{3.4.5...2016}\)

\(A=2015.\dfrac{2}{2016}=2015.\dfrac{1}{1008}=\dfrac{2015}{1008}\)

Vậy \(A=\dfrac{2015}{1008}\)

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Nguyễn Thế Dương
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gãi hộ cái đít
9 tháng 3 2021 lúc 20:30

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

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

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Leonor
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Nguyễn Lê Phước Thịnh
16 tháng 6 2023 lúc 20:29

Sửa đề: A=(1+1/1*3)(1+1/2*4)*...*(1+1/2019*2021)

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

\(=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}=2020\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)

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Leonor
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Ella
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Nguyen My Van
25 tháng 5 2022 lúc 8:56

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

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

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

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

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suli
25 tháng 5 2022 lúc 9:00

undefined

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Nguyễn Lê Phước Thịnh
12 tháng 7 2023 lúc 23:49

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

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

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

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Nguyễn Minh Dương
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Minh Hiếu
17 tháng 10 2023 lúc 14:34

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

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

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

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nguyễn thị hương giang
17 tháng 10 2023 lúc 14:37

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

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

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

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

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

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

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Nguyễn Minh Dương
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Nguyễn Thị Thương Hoài
19 tháng 9 2023 lúc 15:57

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

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

P = \(\dfrac{2021}{1011}\)

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