Tìm số nguyên x, y biết: (x - 1).(y + 2) = 7

\(\left( {x - 1} \right).\left( {y + 2} \right) = 7\)

Ta có:   \[7 = 1.7 = \left( { - 1} \right).\left( { - 7} \right)\]

Vì \[x,y{\rm{ }} \in \mathbb{Z}\;\] nên \[x - 1{\rm{ }} \in \mathbb{Z}\;;\,\,y + 2{\rm{ }} \in \mathbb{Z}\] và \(\left( {x - 1} \right)\left( {y + 2} \right) = 7\)

Suy ra:    +   \[x - 1{\rm{ = 1}}\;;\,\,y + 2{\rm{ = 7}}\]    \[ \Leftrightarrow x{\rm{ = }}\,\,{\rm{2}}\;;\,\,y{\rm{ = }}\,\,{\rm{5}}\]

               +   \[x - 1{\rm{ = 7}}\;;\,\,y + 2{\rm{ = 1}}\]    \[ \Leftrightarrow x{\rm{ = }}\,\,8\;;\,\,y{\rm{ = }}\,\, - 1\]   

               +   \[x - 1{\rm{ = }}\,\,{\rm{ - 1}}\;;\,\,y + 2{\rm{ = }}\,\,{\rm{ - 7}}\] \[ \Leftrightarrow x{\rm{ = }}\,\,0\;;\,\,y{\rm{ = }}\,\, - 9\]  

               +   \[x - 1{\rm{ = }}\,\,{\rm{ - 7}}\;;\,\,y + 2{\rm{ = }}\,\,{\rm{ - 1}}\] \[ \Leftrightarrow x{\rm{ = }}\,\, - 6\;;\,\,y{\rm{ = }}\,\, - 3\]  

Vậy \[\;\left( {x;y} \right) \in \left\{ {\left( {2;5} \right);\left( {8; - 1} \right):\left( {0; - 9} \right);\left( { - 6; - 3} \right)} \right\}\]