Ta có x = 2020

=> x + 1 = 2021

A = x²⁰²¹ - 2021x²⁰²⁰ + .... + 2021x - 2021

= x²⁰²¹ - (x + 1)x²⁰²⁰ + .... + (x + 1)x - (x + 1)

= x²⁰²¹ - x²⁰²¹ - x²⁰²⁰ + .... + x² + x - x + 1

= 1

Vậy A = 1

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

\(A=x^{2021}-\left(x+1\right)x^{2020}+\left(x+1\right)x^{2019}-\left(x+1\right)x^{2018}+...-\left(x+1\right)x^2+\left(x+1\right)x-2021\)

= x²⁰²¹ - x²⁰²¹ - x²⁰²⁰  + x²⁰²⁰ + x²⁰¹⁹ - x²⁰¹⁹ - x²⁰¹⁸ + ... - x³ - x² + x² + x - 2021 = x - 2021 

mà x = 2020 hay 2020 - 2021 = -1 

Vậy với x = 2020 thì A = -1

x = 2020 => 2021 = x + 1

x²⁰²⁰ - 2021x²⁰¹⁹ + 2021x²⁰¹⁸ - 2021x²⁰¹⁷ + ... + 2021x² - 2021x + 1

= x²⁰²⁰ - ( x + 1 )x²⁰¹⁹ + ( x + 1 )x²⁰¹⁸ - ( x + 1 )x²⁰¹⁷ + ... + ( x + 1 )x² - ( x + 1 )x + 1

= x²⁰²⁰ - x²⁰²⁰ - x²⁰¹⁹ + x²⁰¹⁹ + x²⁰¹⁸ - x²⁰¹⁸ - x²⁰¹⁷ + ... + x³ + x² - x² - x + 1

= -x + 1 = -2020 + 1 = -2019

Vậy giá trị của biểu thức = -2019

f(2020) = 2020⁶ - 2021 × 2020⁵ + 2021 × 2020⁴ - 2021×2020³ + 2021×2020² - 2021 × 2020 + 2021 = 1

Chúc bn học tốt !!!!!!!

Vì x = 2020 

=> x + 1 = 2021

Khi đó f(2020) = x⁶ - (x + 1)x⁵ + (x + 1)x⁴ - (x + 1)x³ + (x + 1)x² - (x + 1)x + (x + 1) 

= x⁶ - x⁶ - x⁵ + x⁵ + x⁴ - x⁴ - x³ + x³ + x² - x² - x + x + 1

= 1

Ta có: a = 2020 => 2021 = x + 1

f(2020) = x²⁰¹⁴ - (x + 1) . x²⁰¹³ + (x + 1) . x²⁰¹² - ... + (x + 1) . x² - (x + 1) . x - 1

= x²⁰¹⁴ - x²⁰¹⁴ + x²⁰¹³ + x²⁰¹³ + x²⁰¹² - ... + x³ + x² - x² + x - 1

= x - 1 = 2020 - 1 = 2019

Vậy f(2020) = 2019

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

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

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

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

Ta có: \(\left|x+\frac{1}{2021}\right|\ge0\) ; \(\left|x+\frac{2}{2021}\right|\ge0\) ; ... ; \(\left|x+\frac{2020}{2021}\right|\ge0\) \(\left(\forall x\right)\)

\(\Rightarrow\left|x+\frac{1}{2021}\right|+\left|x+\frac{2}{2021}\right|+...+\left|x+\frac{2020}{2021}\right|\ge0\left(\forall x\right)\)

\(\Rightarrow2021x\ge0\Rightarrow x\ge0\)

Từ đó ta được: \(x+\frac{1}{2021}+x+\frac{2}{2021}+...+x+\frac{2020}{2021}=2021x\)

\(\Leftrightarrow2020x+\frac{1+2+...+2020}{2021}=2021x\)

\(\Leftrightarrow x=\frac{\left(2020+1\right)\left[\left(2020-1\right)\div1+1\right]}{2021}\)

\(\Leftrightarrow x=\frac{2021\cdot2020}{2021}=2020\)

Vậy x = 2020

\(\left|\frac{x+1}{2021}\right|+\left|\frac{x+2}{2021}\right|+...+\left|\frac{x+2020}{2021}\right|=2021x\)

Ta có:\(\left|\frac{x+1}{2021}\right|\ge0;\left|\frac{x+2}{2021}\right|\ge0;....;\left|\frac{x+2020}{2021}\right|\ge0\forall x\)

\(\Rightarrow\left|\frac{x+1}{2021}\right|+\left|\frac{x+2}{2021}\right|+...+\left|\frac{x+2020}{2021}\right|\ge0\forall x\)

\(\Rightarrow2021x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\frac{x+1}{2021}+\frac{x+2}{2021}+...+\frac{x+2020}{2021}=2021x\)

\(\Rightarrow x+\frac{1}{2021}+x+\frac{2}{2021}+...+x+\frac{2020}{2021}=2021x\)

\(\Rightarrow2020x+\frac{1+2+...+2020}{2021}=2021x\)

\(\Rightarrow x=2020\)

\(x=2020\\ \Leftrightarrow x+1=2021\)

Thay vào biểu thức:

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

x=2020

=>x+1=2021

thay vào ta có

\(=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+2021\)

\(=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+2021\)

\(=-x+2021\)

\(=-2020+2021\)

\(=1\)

\(x^6-\left(2020+1\right)x^5+\left(2020+1\right)x^4-\left(2020+1\right)x^3+\left(2020+1\right)x^2-\left(2020+1\right)x+2021\)

\(x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+2021\) (vì x=2020)

\(x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+2021\)\(=-x+2021=2021-2020=1\)

(bạn không hiểu chỗ nào thì hỏi lại mik nhé) chúc bạn buổi sáng tốt lành

Thay 2021 = x + 1 vào A

A = x⁶ - ( x + 1 ) .x⁵ + ( x + 1 ). x⁴  -  ( x + 1 ). x³ + ( x + 1 ) .x² - ( x + 1 ) .x + ( x + 1 )

   = x⁶ - x⁶ - x⁵ + x⁵ + x⁴ - x⁴ - x³ + x³ + x² - x² - x + x + 1

  = 1

Vậy A = 1