a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)

\(=4x^2-20x+25-4x^2+20x\)

=25

b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)

\(=16-9x^2+9x^2+6x+1\)

=6x+17

c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)

\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)

=1

d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)

\(=\left(2021x-2020-2020x+2021\right)^2\)

\(=\left(x+1\right)^2\)

\(=x^2+2x+1\)

f(x) = \(\left(x^6-2019x^5\right)-\left(x^5-2019x^4\right)+\left(x^4-2019x^3\right)-\left(x^3-2019x^2\right)+\left(x^2-2019x\right)-\left(x-2019\right)+1\)

= \(x^5\left(x-2019\right)-x^4\left(x-2019\right)+x^3\left(x-2019\right)-x^2\left(x-2019\right)+x\left(x-2019\right)-\left(x-2019\right)+1\)

Thay x = 2019 vào f(x), ta có:

f(2019) = 0 + 0 + 0 + 0 + 0 +0 + 1 = 1

\(f\left(2019\right)=x^{100}-\left(2019+1\right)x^{99}+\left(2019+1\right)x^{98}-....+\left(2019+1\right)x^2-\left(2019+1\right)x+2000\)

\(=x^{100}-\left(x+1\right)x^{99}+\left(x+1\right)x^{98}-...+\left(x+1\right)x^2-\left(x+1\right)x+2000\)

\(=x^{100}-x^{100}-x^{99}+x^{99}+x^{98}-...+x^3+x^2-x^2-x+2000\)

\(=-x+2000=-2019+2000\)

\(=-19\)

\(x^{2019}-2020x^{2018}+2020x^{2017}-2020x^{2016}+...+2020x-2020\)

\(=x^{2019}-2019x^{2018}-x^{2018}+2019x^{2017}+x^{2017}\)

\(-2019x^{2016}-x^{2016}+...+2019x+x-2020\)

\(=x^{2018}\left(x-2019\right)-x^{2017}\left(x-2019\right)+x^{2016}\left(x-2019\right)\)

\(+...-x\left(x-2019\right)+\left(x-2019\right)-1\)

\(=-1\)

2020.2019^5 = (2019+1).2019^5 = 2019^6+2019^5 làm tương tự với các x còn lại

A= 2019^6 - 2019^6 +.....-2019^2-2019 +2020 = 1 vậy A=1

ta có x = 2019 \(\Rightarrow\)2020 = x+1  

thay 2020 = x+1 vào A ta có

\(A=x^6-\left(x+1\right).x^5+\left(x+1\right).x^4-...-\left(x+1\right).x+2020\)

\(=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+2020\)

\(=-x+2020\)

\(=-2019+2020\)

\(=1\)

vậy A = 1

học tốt !!!