\(a,=x\left(x^2+2\right)\\ b,=2021\left(x+y\right)+x\left(x+y\right)=\left(x+2021\right)\left(x+y\right)\\ c,=49-\left(2x+y\right)^2=\left(7-2x-y\right)\left(7+2x+y\right)\)

a) x³+2x
  = x(x²+2)
b) 2021x+2021y+x²+xy
  = (2021x+2021y)+(x²+xy)
  = 2021(x+y)+x(x+y)
  = (2021+x)(x+y)
c) 49-4x²-4xy-y²  
  = -[(2x)²+2.2x.y+y²] + 7²
  = -(2x-y)²+7²
  = 7²-(2x-y)²
  = (7-2x+y)(7+2x-y)

a

=y(x+5)

b

=(2021+a)(x-y)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

$(x^2+y^2)(1+2021^2)\geq (x+2021y)^2=1$

$\Rightarrow x^2+y^2\geq \frac{1}{1+2021^2}$
Vậy GTNN của $x^2+y^2$ là $\frac{1}{1+2021^2}$

\(x+2021y=1\Rightarrow x=1-2021y\)

\(\Rightarrow x^2+y^2=\left(1-2021y\right)^2+y^2=4084442x^2-4042x+1=\dfrac{1}{4084442}\left(x-\dfrac{2021}{4084442}\right)^2+\dfrac{1}{4084442}\ge\dfrac{1}{4084442}\)

a) Có x = 2020 => x + 1 = 2021. Thay 2021 = x + 1 vào A

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

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

\(A=1\)

b) Có x = -19 => x - 1 = -20 => - ( x - 1 ) = 20. Thay 20 = - ( x - 1) vào B

\(B=x^{10}-\left(x-1\right)x^9-\left(x-1\right)x^8-\left(x-1\right)x^7-...-\left(x-1\right)x^2-\left(x-1\right)x-x+1\)

\(B=x^{10}-x^{10}+x^9-x^9+...+x^2-x^2+x-x+1\)

\(B=1\)

Chúc bạn học tốt!!!

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