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`D = (1-x^2)/(x^2+ 3)`
Để `D` nguên
`->1-x^2 \vdots x^2 +3`
`-> -x^2 + 1 \vdots x^2 + 3`
`-> -x^2 - 3 +4 \vdots x^2 + 3`
`-> - (x^2 + 3) + 4 \vdots x^2 + 3`
Vì `x^2 + 3 \vdots x^2 + 3 -> - (x^2 + 3) \vdots x^2 +3`
`-> 4 \vdots x^2 + 3`
`->x^2 + 3 ∈ Ư (4) = {1;-1;2;-2;4;-4}`
`-> x^2 ∈ {-2; -4; -1 ;-5; 1; -7}`
Vì `x^2 ≥ 0∀x`
`-> x^2 ∈ {1}`
`-> x^2 ∈ {1^2; (-1)^2}`
`-> x ∈ {1;-1}`
Vậy `x ∈ {1;-1}` để `D` nguyên
