Question

Answer 1

b/ 2020xy= x + y (1)

<=> y= \(\dfrac{-x}{1-2020x}\) ĐK: x\(\ne\)\(\dfrac{1}{2020}\)

Do y thuộc Z => \(\dfrac{-x}{1-2020x}\)thuộc Z => -x \(⋮\)(1-2020x)

=> x \(⋮\)(2020x-1) => 2020x\(⋮\)(2020x-1)=>(2020x -1) +1 \(⋮\)(2020x -1)

=>1 \(⋮\)(2020x -1)

=> (2020x -1) \(\inƯ\left(1\right)=\left\{1,-1\right\}\)

\(\left[{}\begin{matrix}2020x-1=1\\2020x-1=-1\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}x=\dfrac{1}{1010}\left(lọai\right)\\x=0\left(tm\right)\end{matrix}\right.\)

Thay x=0 vào (1) => y=0

Vậy (x,y)=(0,0)