Question

Answer 1

Ta có: \(x^2+xy-2020x-2021y-2022=0\)

      \(\Leftrightarrow x\left(x+y+1\right)-2021\left(x+y+1\right)=1\)

      \(\Leftrightarrow\left(x-2021\right)\left(x+y+1\right)=1\)

TH1: \(\left\{{}\begin{matrix}x-2021=1\\x+y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2022\\y=-2022\end{matrix}\right.\left(tm\right)\)

TH2: \(\left\{{}\begin{matrix}x-2021=-1\\x+y+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2020\\y=-2022\end{matrix}\right.\left(tm\right)\)