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Lục Tiểu Ly
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Giải: 1) A=1/1.3+1/3.5+1/5.7+1/7.9+...+1/2017.2019     A=1/2.(2/1.3+2/3.5+2.5.7+2/7.9+...+2/2017.2019)     A=1/2.(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+...+1/2017-1/2019)     A=1/2.(1/1-1/2019)     A=1/2.2018/2019     A=1009/2019 Chúc bạn học tốt!
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Đinh Thị Quỳnh Anh
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Đỗ Tuệ Lâm
17 tháng 2 2022 lúc 15:56

\(S=\dfrac{1}{2}.\left(\dfrac{2}{\sqrt{1.3}}+\dfrac{2}{\sqrt{3.5}}+.......+\dfrac{2}{\sqrt{29.31}}\right)\)

\(S=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}+\dfrac{1}{\sqrt{5}}+.....-\dfrac{1}{\sqrt{29}}+\dfrac{1}{\sqrt{29}}-\dfrac{1}{\sqrt{31}}\right)\)

\(S=\dfrac{1}{2}.\left(1-\dfrac{1}{\sqrt{31}}\right)=\dfrac{1}{2}.\left(\dfrac{31-\sqrt{31}}{31}\right)=\dfrac{31-\sqrt{31}}{62}\)

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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 Thiện Minh
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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

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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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Quách Trần Gia Lạc
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TNA Atula
17 tháng 1 2018 lúc 21:52

2S=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2007.2009}\)

=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{2007}-\dfrac{1}{2009}\)

= 1- \(\dfrac{1}{2009}\)

= \(\dfrac{2008}{2009}\)

=> S=\(\dfrac{1004}{2009}\)

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Huy Phan Đình
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 Mashiro Shiina
11 tháng 7 2017 lúc 9:06

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

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

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

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

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

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

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

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

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

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

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Lê Hương Giang
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