tính
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right).....\left(\dfrac{1}{2021}\right)\)
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\(=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2021}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{2021}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-2020}{2021}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2022}{2021}\)
\(=\dfrac{1}{2021}\cdot\dfrac{2022}{2}=\dfrac{1011}{2021}\)
-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
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 \(B-A=\dfrac{3^{2013}}{2^{2014}}-\dfrac{3^{2013}}{2^{2012}}+\dfrac{5}{2}\)
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
\(a,2021-\left(\dfrac{1}{3}\right)^2\cdot3^2\\ =2021-\dfrac{1}{9}\cdot9\\ =2021-\dfrac{9}{9}\\ =2021-1=2020\\ b,\dfrac{5}{10}+9\cdot\dfrac{-3}{2}\\ =\dfrac{5}{10}+\dfrac{-27}{2}\\ =\dfrac{5}{10}+\dfrac{-135}{10}\\ =-\dfrac{130}{10}\\ =-13\\ c,-10\cdot\left(-\dfrac{2021}{2022}\right)^0+\left(\dfrac{2}{5}\right)^2:2\\ =-10\cdot1+\dfrac{4}{25}\cdot\dfrac{1}{2}\\ =-10+\dfrac{4}{50}\\ =-10+\dfrac{2}{25}\\ =-\dfrac{248}{25}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2021.2023}\right)\)
\(=\dfrac{4}{1.3}.\dfrac{9}{2.4}...\dfrac{4088484}{2021.2023}\)
\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{2022.2022}{2021.2023}\)
\(=\dfrac{2.2022}{1.2023}\)
a: =2+6*(-1)^2019+2026
=2028-6
=2022
b: \(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}...\cdot\dfrac{625}{624}\)
\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{4^2}{\left(4-1\right)\left(4+1\right)}...\cdot\dfrac{625}{\left(25-1\right)\left(25+1\right)}\)
\(=\dfrac{2\cdot3\cdot4\cdot...\cdot49}{1\cdot2\cdot3\cdot...\cdot48}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot49}{3\cdot4\cdot5\cdot...\cdot50}\)
\(=\dfrac{49}{1}\cdot\dfrac{2}{50}=\dfrac{98}{50}=\dfrac{49}{25}\)
`A=\(\frac{1}{2}-1)(\dfrac{1}{3}-1).....(\frac{1}{2021}-1)`
`=\frac{-1}{2}.\frac{-2}{3}.....\frac{-2020}{2021}`
`=(-1xx(-2)xx...xx(-2020))/(2xx3xx....xx2021)`
`=(1xx2xx..xx2020)/(2xx3xx...xx2021)`(do `1->2020` có 2020 số)
`=\frac{1}{2021}`
Cái cuối có -1 ko