x=2020; x=1

\(x^2+2019x=2020\)
\(x\left(x+2019\right)=2020\)
Tách 2020 ra 2 thừa số có hiệu là 2019: 2020 = 1*2020 = (-1) * (-2020)
Mà thừa số x luôn bé hơn thừa số x + 2019
\(\Rightarrow x\in\left\{1;-2020\right\}\)

 

Các bạn giúp mk vs nhé!

a) \(ĐKXĐ:x\ne\pm3\)

      \(A=\frac{5}{x+3}-\frac{2}{3-x}+\frac{3x^2-2x-9}{x^2-9}\)

\(\Leftrightarrow A=\frac{5\left(x-3\right)+2\left(x+3\right)-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{-3x}{x+3}\)

b) Khi \(\left|x-2\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\2-x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)

Thay x = 1 vào A, ta được :

\(A=\frac{-3}{1+3}=\frac{-3}{4}\)

Vậy khi \(\left|x-2\right|=1\Leftrightarrow A=-\frac{3}{4}\)

c) Để \(A\inℤ\)

\(\Leftrightarrow\frac{-3x}{x+3}\inℤ\)

\(\Leftrightarrow-3x⋮x+3\)

\(\Leftrightarrow-3\left(x+3\right)+9⋮x+3\)

\(\Leftrightarrow9⋮x+3\)

\(\Leftrightarrow x+3\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)

\(\Leftrightarrow x\in\left\{-2;-4;0;-6;-12;6\right\}\)

Vậy để \(A\inℤ\Leftrightarrow x\in\left\{-2;-4;0;-6;-12;6\right\}\)