Dễ thấy \(VT\ge0\Rightarrow2020x\ge0\Leftrightarrow x\ge0\)

\(\Rightarrow pt\Leftrightarrow2019x+\frac{2019.2020}{2}=2020x\)

\(\Leftrightarrow x=2019.1010\)

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

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 

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 !!!

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

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

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

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

\(\left|2020x^2+4040x\right|=\left|x+2\right|\)

\(\Leftrightarrow\left|2020x\left(x+2\right)\right|=\left|x+2\right|\)

\(\Leftrightarrow\left|x+2\right|\left|2020x\right|=\left|x+2\right|\)

\(\Leftrightarrow\left|x+2\right|\left|2020x\right|-\left|x+2\right|=0\)

\(\Leftrightarrow\left|x+2\right|\left(\left|2020x\right|-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left|x+2\right|=0\\\left|2020x\right|-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x\in\left\{\frac{1}{2020};\frac{-1}{2020}\right\}\end{cases}}\)

2021 - x + 2021(x - 2020x) = 0

<=> 2021 - x + 2021 - 4082420 = 0

<=> -x - 4082420 = 0

<=> x = -4082420

a: \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}B=\dfrac{-3}{3-1}=\dfrac{-3}{2}\\B=\dfrac{1}{-1-1}=-\dfrac{1}{2}\end{matrix}\right.\)