$1,\\a+b=c+\dfrac{1}{2021}\\<=> \dfrac{1}{2021}=a+b-c\\<=> 2021=\dfrac{1}{a+b-c}\\\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{c}+2021\\<=>\dfrac{a+b}{ab}=\dfrac{1}{c}+\dfrac{1}{a+b-c}\\<=>\dfrac{a+b}{ab}=\dfrac{a+b-c+c}{c(a+b-c)}\\<=>\dfrac{a+b}{ab}=\dfrac{a+b}{c(a+b-c)}\\<=> (a+b)(\dfrac{1}{ab}-\dfrac{1}{c(a+b-c)})=0\\<=>\dfrac{(a+b)(c(a+b-c)-ab)}{abc(a+b-c)}=0\\<=>(a+b)(ac+bc-c^2-ab)=0\\<=>(a+b)(c(a-c)-b(a-c))=0\\<=>(a+b)(c-b)(a-c)=0$

TH1 : $a+b=0<=>a=-b$

$=>A=-c^{2021} . (-\dfrac{1}{c^{2021}})=1$

TH2 : $b=c$

$=> A=a^{2021}.\dfrac{1}{a^{2021}}=1$

TH3 : $a=c$

$=>A=b^{2021}.\dfrac{1}{b^{2021}}=1$

Vậy $A=1$

$2,$

Đặt $x^2+1=t$

$<=>t^2+3xt +2x^2=0\\<=>t^2+2xt+tx+2x^2=0\\<=> t(t+2x)+x(t+2x)=0\\<=>(t+x)(t+2x)=0$

TH1 : $t=-x$

$<=> x^2+x+1=0$ (Vô nghiệm)

TH2 : $t=-2x$

$<=>x^2+2x+1=0\\<=>(x+1)^2=0\\<=>x=-1$

Vậy pt có nghiệm duy nhất $x=-1$

`2)`

   `(x^2+1)^2+3x(x^2+1)+2x^2=0`

`⇔x^4+2x^2+1+3x^3+3x+2x^2=0`

`⇔x^4+3x^3+4x^2+3x+1=0`

`⇔x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0`

`⇔x^3(x+1)+2x^2(x+1)+2x(x+1)+(x+1)=0`

`⇔(x+1)(x^3+2x^2+2x+1)=0`

`⇔(x+1)(x^3+x^2+x^2+x+x+1)=0`

`⇔(x+1)[x^2(x+1)+x(x+1)+(x+1)]=0`

`⇔(x+1)(x+1)(x^2+x+1)=0`

`⇔(x+1)^2(x^2+2x.1/2+1/4+3/4)=0`

`⇔(x+1)^2[(x+1/2)^2+3/4]=0`

  Mà `(x+1/2)^2+3/4 > 0`

 `⇒(x+1)^2=0`

`⇔x+1=0`

`⇔x=-1`

Vậy ptr đã cho có nghiệm `x=-1`