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

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 !!!

\(D=4x^2-2x+3x\left(x-5\right)=4x^2-2x+3x^2-15x=7x^2-17x=7\left(-1\right)^2-17\left(-1\right)=24\)

\(E=x^{10}-2020x^9+2020x^8-2020x^7+...+2020x^2-2020x=x^9\left(x-2019\right)-x^8\left(x-2019\right)+x^7\left(x-2019\right)-...-x^2\left(x-2019\right)+x\left(x-2019\right)-x=x^9\left(2019-2019\right)-...+x\left(2019-2019\right)-2019=-2019\)

Bài làm:

Ta có: \(x=2019\Rightarrow2020=x+1\)

Thay vào ta được:

\(f\left(2019\right)=x^{99}-\left(x+1\right)x^{98}+\left(x+1\right)x^{97}-\left(x+1\right)x^{96}+...-\left(x+1\right)x^2+\left(x+1\right)x-1\)

\(f\left(2019\right)=x^{99}-x^{99}-x^{98}+x^{98}+x^{97}-x^{97}-x^{96}+...-x^3-x^2+x^2+x-1\)

\(f\left(2019\right)=x-1\)

Thay \(x=2019\)vào ta được:

\(f\left(2019\right)=2019-1=2018\)

Vậy f(2019) = 2018

\(f\left(x\right)=x^{99}-2020x^{98}+2020x^{97}-2020x^{96}+...-2020x^2+2020x-1\)

\(f\left(2019\right)=2019^{99}-2020.2019^{98}+2020.2019^{97}-...+2020.2019-1\)

Xét  \(2020.2019^{98}=2019^{99}+2019^{98};2020.2019^{97}=2019^{98}+2019^{97}\)

\(2020.2019^{96}=2019^{97}+2019^{96};...;2020.2019=2019^2+2019\)

\(\Rightarrow f\left(2019\right)=2019^{99}-2019^{99}-2019^{98}+2019^{97}-2019^{97}-...+2019^2+2019-1\)

\(\Rightarrow f\left(2019\right)=2019-1=2018\). Vậy \(f\left(2019\right)=2018\)

Ta có: \(x=2021\Rightarrow2020=x-1\)

Thay vào được:

\(A=x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x\)

\(A=x^4-x^4+x^3-x^3+x^2-x^2+x\)

\(A=x=2021\)

Vậy A  = 2021

Ta có: \(x=2021\)\(\Rightarrow x-1=2020\)

Thay \(x-1=2020\)vào biểu thức A ta được:

\(A=x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x\)

\(=x^4-x^4+x^3-x^3+x^2-x^2+x\)

\(=x=2021\)

Theo bài ra ta có : \(x=2021\Leftrightarrow x-1=2020\)

hay \(A=x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x\)

\(=x^4-x^4+x^3-x^3+x^2-x^2+x=x\)

\(\Rightarrow x=2021\)Vậy biểu thức A nhận giá trị 2021 

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

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

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

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

\(=-19\)