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

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

Theo kinh nghiệm của tui thì.......mấy cái bài này hay dễ ra kết quả = 1 với = 0 nhiều lắm:)

\(S=\left(-\dfrac{1}{5}\right)^0+\left(-\dfrac{1}{5}\right)^1+...+\left(-\dfrac{1}{5}\right)^{2021}\)

\(\Rightarrow\left(-\dfrac{1}{5}\right)S=\left(-\dfrac{1}{5}\right)^1+\left(-\dfrac{1}{5}\right)^2+...+\left(-\dfrac{1}{5}\right)^{2021}+\left(-\dfrac{1}{5}\right)^{2022}\)

\(\Rightarrow S-\left(-\dfrac{1}{5}\right)S=\left(-\dfrac{1}{5}\right)^0-\left(-\dfrac{1}{5}\right)^{2022}\)

\(\Rightarrow\dfrac{6}{5}S=1-\dfrac{1}{5^{2022}}\)

\(\Rightarrow S=\dfrac{5}{6}\left(1-\dfrac{1}{5^{2022}}\right)\)

\(\dfrac{-5}{2}-\left(\dfrac{-7}{2}-\dfrac{7}{4}\right):\left(\dfrac{-7}{3}\right)-\left(\dfrac{-7}{3}\right)-2021^0\)

\(=\dfrac{-5}{2}-\dfrac{9}{4}+\dfrac{7}{3}-1\)

\(=-\dfrac{41}{12}\)

1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25

làm ơn giúp 🙏🙏🙏

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9

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

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

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

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