Question

tìm các cặp số nguyên (x;y) thỏa mãn: xy-2y+x-3=0

Answer 1

\(\Leftrightarrow y\left(x-2\right)+\left(x-2\right)-1=0\)

\(\Leftrightarrow\left(x-2\right)\left(y+1\right)=1\)

TH1:

\(\left\{{}\begin{matrix}x-2=1\\y+1=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}x-2=-1\\y+1=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy (x;y) = (3;0); ( 1;-2)