Ta có : `(x.x+3x-7)/(x+3)`
⇒ $\left \{ {{x.x+3.x-7÷x+3} \atop {x+3÷x+3}} \right.$
⇒ $\left \{ {{x.x+3.x-7÷x+3} \atop {x.x+3.x÷x+3}} \right.$
`⇒ (x.x+3.x)-(x.x+3.x-7)÷ x+3`
`⇔ 7÷x+3`
`⇒ x+3∈ Ư(7)`
`⇒ x+3∈{1;7;-1;-7}`
`⇔ x∈{ -2;4;-4;-10}`
Vậy để `(x.x+3x-7)/(x+3)` nguyên thì ` x∈{ -2;4;-4;-10}`