Đặt 20212020=x

=>\(A=\dfrac{3\left(x+1\right)\left(x+3\right)-5x-2\cdot\left(x+1\right)^2-5}{\left(x+1\right)}\)

\(=\dfrac{3\left(x^2+4x+3\right)-5x-2x^2-4x-2-5}{\left(x+1\right)}\)

\(=\dfrac{3x^2+12x+9-2x^2-9x-7}{x+1}=\dfrac{x^2+3x+2}{x+1}=x+2\)

=20212022

2021²⁰²²

2021^ 2022

-2021 . 20222022 + 2022 .20212021

= -2021 . 2022. 10001 + 2022.2021. 10001

= 0

-2021 . 20222022 + 2022 . 20212021

= (-2021) . (20222022 - 2022 . 10001)

= (-2021) . (20222022 - 20222022)

= (-2021) . 0

= 0

2021

2021¹ = 2021

=2021

\(\frac{20212021}{20222022}=\frac{20212021:10001}{20222022:10001}=\frac{2021}{2022}\)

   \(\dfrac{2021}{2022}\) x \(\dfrac{2022020222022}{202320232023}\) x \(\dfrac{20212021}{20232023}\)

= \(\dfrac{2021}{2022}\) x \(\dfrac{2022}{2023}\) x \(\dfrac{2021}{2023}\)

= \(\dfrac{2021\times2021}{2023\times2023}\)

= \(\dfrac{4084441}{4092529}\)

dư 123

123

Dư 123

   \(\dfrac{121212}{242424}+\dfrac{1313}{2626}+\dfrac{20212021}{40424042}\)

= \(\dfrac{12}{24}+\dfrac{13}{26}+\dfrac{2021}{4042}\)

= \(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\)

= \(\dfrac{3}{2}\)

121212/242424+1313/2626+20212021/404244024

= 1/2+1/2+1/2 

=3/2

121212/242424 + 1313/2626 + 20212021/40424042

= 1/2 + 1/2 + 1/2

= 3/2