a: \(2022^{x-2021}+3=\left(7-5\right)^2\)

=>\(2022^{x-2021}+3=4\)

=>\(2022^{x-2021}=1\)

=>x-2021=0

=>x=2021

b: \(\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=795\)

=>\(30x+\left(1+2+3+...+30\right)=795\)

=>\(30x+\dfrac{30\cdot31}{2}=795\)

=>\(30x=795-31\cdot15=330\)

=>x=11

Khi x = 2021

=> 2022 = x + 1

Khi đó E = x¹⁰ - 2022x⁹ + 2022x⁸ - ... + 2022x² - 2022x + 2022

= x¹⁰ - (x + 1)x⁹ + (x + 1)x⁸ - .... + (x + 1)x² - (x + 1)x + (x + 1) 

= x¹⁰ - x¹⁰ - x⁹ + x⁹ + x⁸ - ... + x³ + x² - x² - x + x + 1

= 1 

\(\Rightarrow2022^{x-2021}=2^2-3=1=2022^0\\ \Rightarrow x-2021=0\\ \Rightarrow x=2021\)

2022^x-2021+ 3 = 4

2022^x-2021      = 4 - 3

2022^x-2021      = 1

➩x - 2021 = 0 

    x            = 0 + 2021

    x            = 2021

Vậy x = 2021

\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\Rightarrow x=-2\\ b,\Rightarrow\left(x-2021\right)\left(2022x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{2022}\end{matrix}\right.\\ c,\Rightarrow\left(x^2-9\right)-\left(x-3\right)\left(2x+7\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(2x+7\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3-2x-7\right)=0\\ \Rightarrow\left(x-3\right)\left(-4-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

\(a,Sửa:2021x-1+2022x\left(1-2021x\right)=0\\ \Leftrightarrow\left(2021x-1\right)\left(1-2022x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2021}\\x=\dfrac{1}{2022}\end{matrix}\right.\)

\(Q\left(x\right)=x^{101}-2020x^{100}-2022x^{99}+2022x^{98}+x-2021\)

\(=x^{100}\left(x-2021\right)+x^{99}\left(x-2021\right)-x^{98}\left(x-2021\right)+x^{98}+x-2021\)

\(Q\left(2021\right)=0+0-0+2021^{98}+0=2021^{98}\)

Thay `x=2021` vào A: `A=2020.2021-2022 .2021^2 +2021^3=-2021`

x=2021⇒2020=x-1; 2022=x+1, thay vào A ta có:

A=2020x-2022x²+x³

=(x-1)x-(x+1)x²+x³

=x²-x-x³-x²+x³

=x

=2021