Tìm số nguyên x, y biết: x(y - 3) = -6

\(x\left( {y - 3} \right) = - 6\)               

Ta có:   \[ - 6\; = \left( { - 1} \right).6 = 1.\left( { - 6} \right) = \left( { - 2} \right).3 = 2.\left( { - 3} \right)\]

Vì \[x,y{\rm{ }} \in \mathbb{Z}\;\] nên \[y - 3{\rm{ }} \in \mathbb{Z}\;\]và \(x\left( {y - 3} \right) = - 6\)

Suy ra: + \(x = \,\, - 1\,\,;y - 3 = 6\)\( \Leftrightarrow x = \,\, - 1\,\,;y = 9\)

             + \(x = \,\,6\,\,;y - 3 = - 1\)\( \Leftrightarrow x = \,\,6\,\,;y = 2\)

             + \(x = \,\,1\,\,;y - 3 = - 6\)\( \Leftrightarrow x = \,\, - 1\,\,;y = - 3\)

             + \(x = \,\, - 6\,\,;y - 3 = 1\)\( \Leftrightarrow x = \,\, - 6\,\,;y = 4\)

             + \(x = \,\,2\,\,;y - 3 = - 3\)\( \Leftrightarrow x = \,\,2\,\,;y = 0\)

             + \(x = \,\, - 3\,\,;y - 3 = 2\)\( \Leftrightarrow x = \,\, - 3\,\,;y = 5\)

             + \(x = \,\,3\,\,;y - 3 = - 2\)\( \Leftrightarrow x = \,\,3\,\,;y = 1\)

             + \(x = \,\, - 2\,\,;y - 3 = 3\)\( \Leftrightarrow x = \,\, - 2\,\,;y = 6\)

Vậy \[\;\left( {x;y} \right) \in \left\{ {\left( { - 1;9} \right);\left( {6;2} \right):\left( { - 1; - 3} \right);\left( { - 6;4} \right);\left( {2;0} \right);\left( { - 3;5} \right);\left( {3;1} \right);\left( { - 2;6} \right)} \right\}\]