\(2021x\left(x-2020\right)-x+2020=0\)

\(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)

Ta có: \(2021x\left(x-2020\right)-x+2020=0\)

\(\Leftrightarrow\left(x-2020\right)\left(2021x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)

a) Ta có 3 trường hợp :

b) x² > 0 vì :

Khi x là các số nguyên khác 0 thì suy ra x phải là số nguyên dương hoặc nguyên âm. Mà phần lũy thừa của x là số chẵn nên x² chắc chắn lớn hơn 0

\(3x\left(x-2020\right)-x+2020=0\)

\(3x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\left(3x-1\right)\left(x-2020\right)=0\)

\(\orbr{\begin{cases}x=\frac{1}{3}\left(TM\right)\\x=2020\left(TM\right)\end{cases}}\)

\(b,4-9x^2=0\)

\(2^2-\left(3x\right)^2=0\)

\(\left(2-3x\right)\left(2+3x\right)=0\)

\(\orbr{\begin{cases}2-3x=0\\2+3x=0\end{cases}\orbr{\begin{cases}x=\frac{2}{3}\left(TM\right)\\x=-\frac{2}{3}\left(TM\right)\end{cases}}}\)

\(c,x^2-x+\frac{1}{4}=0\)

\(x^2-x+\left(\frac{1}{2}\right)^2=0\)

\(\left(x-\frac{1}{2}\right)^2=0\)

\(x-\frac{1}{2}=0\)

\(x=\frac{1}{2}\)

\(d,x\left(x-3\right)+\left(x-3\right)=0\)

\(\left(x-3\right)\left(x+1\right)=0\)

\(\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\orbr{\begin{cases}x=3\left(TM\right)\\x=-1\left(TM\right)\end{cases}}}\)

\(e,9x\left(x-7\right)-x+7=0\)

\(9x\left(x-7\right)-\left(x-7\right)=0\)

\(\left(9x-1\right)\left(x-7\right)=0\)

\(\orbr{\begin{cases}9x-1=0\\x-7=0\end{cases}\orbr{\begin{cases}x=\frac{1}{9}\left(TM\right)\\x=7\left(TM\right)\end{cases}}}\)

a) 3x(x - 2020) - x + 2020 = 0 

<=> 3x(x - 2020) - (x - 2020) = 0

<=> (3x - 1)(x - 2020) = 0

<=> \(\orbr{\begin{cases}3x-1=0\\x-2020=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2020\end{cases}}\)

Vậy tập nghiệm phương trình là \(S=\left\{\frac{1}{3};2020\right\}\)

b) \(4-9x^2=0\)

<=> \(\left(2-3x\right)\left(2+3x\right)=0\)

<=> \(\orbr{\begin{cases}2-3x=0\\2+3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{2}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{2}{3};-\frac{2}{3}\right\}\)là nghiệm phương trình 

c) \(x^2-x+\frac{1}{4}=0\)

<=> \(\left(x-\frac{1}{2}\right)^2=0\)

<=> \(x-\frac{1}{2}=0\)

<=> \(x=\frac{1}{2}\)

d) x(x - 3) + (x - 3) = 0

<=> (x + 1)(x - 3) = 0

<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

Vậy \(x\in\left\{-1;3\right\}\)là nghiệm phương trình

e) 9x(x - 7) - x + 7 = 0

<=> (9x - 1)(x - 7) = 0

<=> \(\orbr{\begin{cases}9x-1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{9}\\x=7\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{9};7\right\}\)là nghiệm phương trình

1, \(3x\left(x-2020\right)-x+2020=0\)

\(\Leftrightarrow\)  \(3x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)\left(3x-1\right)=0\)  

  \(\Leftrightarrow\)\(\orbr{\begin{cases}x-2020=0\\3x-1=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2020\\3x=1\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2020\\x=\frac{1}{3}\end{cases}}\)

Vậy phương trình có nghiệm x=2020 hoặc x=\(\frac{1}{3}\)

2, \(4-9x^2=0\)

\(\Leftrightarrow4=9x^2\)

\(\Leftrightarrow\frac{4}{9}=x^2\)

\(\Leftrightarrow x=\pm\frac{2}{3}\)

Vậy phương trình có nghiệm x=\(\pm\frac{2}{3}\)

3, \(x^2-x+\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy phương trình có nghiệm x=\(\frac{1}{2}\)

4, \(x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Vậy phương trình có nghiệm x=3 hoặc x= -1

5, \(9x\left(x-7\right)-x+7=0\)

\(\Leftrightarrow9x\left(x-7\right)-\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(9x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\9x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\9x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=\frac{1}{9}\end{cases}}\)

Vậy phương trình có nghiệm x=7 hoặc x=\(\frac{1}{9}\)

(x-2020)^{x - 1} - (x - 2020)^{x + 2019} = 0

=> (x - 2020)^{x - 1} .[(x - 2020)²⁰²⁰ - 1] = 0 

=> \(\orbr{\begin{cases}\left(x-2020\right)^{x-1}=0\\\left(x-2020\right)^{2020}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x-2020=0\\\left(x-2020\right)^{2020}=1^{2020}\end{cases}\Rightarrow}\orbr{\begin{cases}x-2020=0\\x-2020=\pm1\end{cases}}}\)

=> \(x-2020\in\left\{0;1;-1\right\}\Rightarrow x\in\left\{2020;2021;2019\right\}\)

(x+2019)(x-2020)=0.

=> x+2019=0 hoặc x-2020=0.

+, x+2019=0.                      +, x-2020=0

       x= 0-2019                         x = 0+2020

      x = -2019.                         x = 2020.

Vậy: x thuộc{ -2019 ; 2020 }.

#Học tốt.

\(\Leftrightarrow\orbr{\begin{cases}x+2019=0\\x-2020=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2019\\x=2020\end{cases}}}\)

\(\left(x+2019\right)\left(x-2020\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2019=0\\x-2020=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-2019\\x=2020\end{cases}}\)

\(x+y=2\Rightarrow y=2-x\)

\(P=2x^2-\left(2-x\right)^2-5x+\dfrac{1}{x}+2020=x^2-x+\dfrac{1}{x}+2016\)

\(P=x^2+1-x+\dfrac{1}{x}+2015\ge2x-x+\dfrac{1}{x}+2015\)

\(P\ge x+\dfrac{1}{x}+2015\ge2\sqrt{\dfrac{x}{x}}+2015=2017\)

Dấu "=" xảy ra khi \(x=y=1\)

    \(x+y+z=0\)

\(\Leftrightarrow\)\(\left(x+y+z\right)^2=0\)

\(\Leftrightarrow\)\(x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)

\(\Leftrightarrow\)\(x^2+y^2+z^2=0\)   (vì  xy + yz + xz = 0)

\(\Rightarrow\)\(x=y=z=0\)

Vậy   \(Q=\left(x-1\right)^{2018}+\left(y-1\right)^{2019}+\left(z-1\right)^{2020}=1\)