tìm số nguyên x để x^2+3x-5 chia hết cho x+2

`(x^2 + 3x - 5)/(x+2)`     (`đkxđ: x \ne 2`)

`= (x^2 + 3x +2 - 7)/(x+2)`

`= (x^2 + x + 2x + 2)/(x+2) - 7/(x+2)`

`= ((x+1)(x+2))/(x+2) - 7/(x+2)`

Để `x^2 + 3x - 5 \vdots x+2`

`⇒ 7 \vdots x+2`

`⇒ x+2 ∈ Ư(7)={±1;±7}`

`⇒ x ∈ {-3;-1;5;-9}`

Vậy,..

`color{Purple}{@Mon}`