c,
$\dfrac{3x +2}{x - 2} = \dfrac{3x - 6 +8 }{x - 2} = \dfrac{3 ( x - 2 ) + 8}{x - 2} = \dfrac{3 ( x - 2 )}{x - 2} + \dfrac{8}{x - 2} = 1 + \dfrac{8}{x - 2}$
Để : $\dfrac{3x +2}{x - 2}$ có giá trị nguyên thì $8 \vdots x - 2$
$=> x - 2 \in Ư ( 8 ) \in$ { $1 ; -1 ; 2 ; -2 ; 4 ; -4 ; 8 ; -8$ }
$x - 2 = 1 -> x = 3$
$x - 2 = -1 -> x = -1$
$x - 2 = 2 -> x = 4$
$x - 2 = -2 -> x = 0$
$x - 2 = 4 -> x = 6$
$x - 2 = -4 -> x = -2$
$x - 2 = 8 -> x = 10$
$x - 2 = -8 -> x = -6$
$x \in$ { $3 ; -1 ; 4 ; 0 ; 6 ; -2 ; 10 ; -6$ }
