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