xy - x + 2y = 3
=> ( xy - x ) + 2y = 3
=> x ( y - 1 ) + 2 ( y - 1 ) + 2 = 3
=> ( y - 1 ) . ( x + 2 ) = 3 - 2
=> ( y - 1 ) . ( x + 2 ) = 1 = 1 . 1 = ( -1 ) . ( -1 )
TH1 :
\(\hept{\begin{cases}y-1=1\\x+2=1\end{cases}}\)=> \(\hept{\begin{cases}y=2\\x=-1\end{cases}}\)
TH2 :
\(\hept{\begin{cases}y-1=-1\\x+2=-1\end{cases}}\)=> \(\hept{\begin{cases}y=0\\x=-3\end{cases}}\)
Vậy : ( x ; y ) \(\in\){ ( -1 ; 2 ) ; ( -3 ; 0 }
Chúc bn học vui^^