a) A có giá trị nguyên `⇔ 3 vdots x-1`
`⇔ x-1 ∈ Ư(3) = {1;-1;3;-3}`
`⇔ x ∈ {2;0;4;-2}`
Vậy `x = 2;0;4;-2`
b) B có giá trị nguyên `⇔ x-2 vdots x+3`
`⇔ x - 2 - (x+3) vdots x+3`
`⇔ x - 2 - x - 3 vdots x+3`
`⇔ -5 vdots x+3`
`⇔ x+3 ∈ Ư(-5) = {1;-1;5;-5}`
`⇔ x ∈ {-2;-4;2;-8}`
Vậy `x = -2;-4;2;-8`
c) C có giá trị nguyên `⇔ 2x+1 vdots x-3`
Ta có: `x - 3 vdots x - 3 => 2x - 6 vdots x - 3`
`⇔ 2x + 1 - (2x - 6) vdots x - 3`
⇔ 2x + 1 - 2x + 6 vdots x - 3`
`⇔ 7 vdots x - 3`
`⇔ x-3 ∈ Ư(7) = {1;-1;7;-7}`
`⇔ x ∈ {4;2;10;-4}`
Vậy `x = 4;2;10;-4`
d) D có giá trị nguyên `⇔ x^2 - 1 vdots x+1`
Ta có: `x + 1 vdots x+1 => x^2 + x vdots x+1`
`⇔ x^2 + x - ( x^2 - 1) vdots x+1`
`⇔ x^2 + x - x^2 + 1 vdots x+1`
`⇔ x + 1 vdots x+1`
`⇔ x ∈Z`
Vậy `x ∈ Z`
(Chúc bạn học tốt)
