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Những câu hỏi liên quan
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Nguyễn Lê Phước Thịnh
2 tháng 7 2023 lúc 22:24

\(=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2021}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{2021}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-2020}{2021}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2022}{2021}\)

\(=\dfrac{1}{2021}\cdot\dfrac{2022}{2}=\dfrac{1011}{2021}\)

Nguyễn Việt Bách
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Đào Thị Nguyên Minh
30 tháng 12 2023 lúc 22:05

-3/4 . -8/9 . ... . -4084440/4084440

= 3/4 . 8/9 . 4084440/4084441

=1.3/2.2  .  2.4/3.3  ...  2020.2022/2021.2021

=1.3.2.4...2020.2022/2.2.3.3...2021.2021

=(1.2...2020)(3.4...2022)/(2.3...2021)(2.3...2021)

=1.2022/2021.2=2022/4042

ko bt
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My name is Chicky
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Nguyen My Van
19 tháng 5 2022 lúc 16:33

Ta có \(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2021}\left(1\right)\)

\(\Rightarrow\dfrac{3}{2}A=\dfrac{3}{4}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2013}\left(2\right)\)

Lấy (2) - (1) ta được:

\(\dfrac{3}{2}A-A=\left(\dfrac{3}{2}\right)^{2013}+\dfrac{3}{4}-\dfrac{1}{2}-\dfrac{3}{2}\)

\(\dfrac{1}{2}A=\left(\dfrac{3}{2}\right)^{2013}+\dfrac{1}{4}\Rightarrow A=\dfrac{3^{2013}}{2^{2012}}+\dfrac{1}{2}\)

Vậy \(B-A=\dfrac{3^{2013}}{2^{2014}}-\dfrac{3^{2013}}{2^{2012}}+\dfrac{5}{2}\)

Lê Phương Linh
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Kiều Vũ Linh
31 tháng 10 2023 lúc 11:55

a) 2021 - (1/3)² . 3²

= 2021 - 1/9 . 9

= 2021 - 1

= 2020

b) 5/10 + 9 . (-3/2)

= 1/2 - 27/2

= -26/2

= -13

c) -10 . (-2021/2022)⁰ + (2/5)² : 2

= -10 . 1 + 4/25 . 2

= -10 + 8/25

= -68/7

⭐Hannie⭐
31 tháng 10 2023 lúc 11:55

\(a,2021-\left(\dfrac{1}{3}\right)^2\cdot3^2\\ =2021-\dfrac{1}{9}\cdot9\\ =2021-\dfrac{9}{9}\\ =2021-1=2020\\ b,\dfrac{5}{10}+9\cdot\dfrac{-3}{2}\\ =\dfrac{5}{10}+\dfrac{-27}{2}\\ =\dfrac{5}{10}+\dfrac{-135}{10}\\ =-\dfrac{130}{10}\\ =-13\\ c,-10\cdot\left(-\dfrac{2021}{2022}\right)^0+\left(\dfrac{2}{5}\right)^2:2\\ =-10\cdot1+\dfrac{4}{25}\cdot\dfrac{1}{2}\\ =-10+\dfrac{4}{50}\\ =-10+\dfrac{2}{25}\\ =-\dfrac{248}{25}\)

Vương Thanh Huyền
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Dương Ngọc Ánh
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alibaba nguyễn
3 tháng 12 2023 lúc 20:25

\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2021.2023}\right)\)

\(=\dfrac{4}{1.3}.\dfrac{9}{2.4}...\dfrac{4088484}{2021.2023}\)

\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{2022.2022}{2021.2023}\)

\(=\dfrac{2.2022}{1.2023}\)

Vũ Ngọc Diệp
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Nguyễn Lê Phước Thịnh
25 tháng 2 2023 lúc 14:09

a: =2+6*(-1)^2019+2026

=2028-6

=2022

b: \(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}...\cdot\dfrac{625}{624}\)

\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{4^2}{\left(4-1\right)\left(4+1\right)}...\cdot\dfrac{625}{\left(25-1\right)\left(25+1\right)}\)

\(=\dfrac{2\cdot3\cdot4\cdot...\cdot49}{1\cdot2\cdot3\cdot...\cdot48}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot49}{3\cdot4\cdot5\cdot...\cdot50}\)

\(=\dfrac{49}{1}\cdot\dfrac{2}{50}=\dfrac{98}{50}=\dfrac{49}{25}\)

Đại Mỹ Trần
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ILoveMath
5 tháng 12 2021 lúc 8:52

\(\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{3}\right)^2+2\left(-\dfrac{1}{2}\right)^2-2021^0\\ =\dfrac{1}{32}:\dfrac{1}{9}+2.\dfrac{1}{4}-1\\ =\dfrac{9}{32}+\dfrac{1}{2}-1\\ =-\dfrac{7}{32}\)

Rin•Jinツ
5 tháng 12 2021 lúc 8:52

\(-\dfrac{7}{32}\)