*Bổ sung đề
$B=\dfrac{2x}{x+3}+\dfrac{2}{x-3}+\dfrac{x^2-x+6}{9-x^2}$ `(x\ne±3)`
$=\dfrac{2x(x-3)}{(x+3)(x-3)}+\dfrac{2(x+3)}{(x-3)(x+3)}-\dfrac{x^2-x+6}{x^2-9}$
$=\dfrac{2x^2-6x+2x+6-x^2+x-6}{(x-3)(x+3)}$
$=\dfrac{x^2-3x}{(x-3)(x+3)}$
$=\dfrac{x(x-3)}{(x-3)(x+3)}$
$=\dfrac{x}{x+3}$
$c)\dfrac{x}{x+3}$
$=\dfrac{x+3}{x+3}-\dfrac{3}{x+3}$
Đề ptrình nguyên thì `3\vdotsx+3`
`Ư(3)∈{±1,±3}`
`->x+3=-1⇔x=-4`(thỏa mãn)
`->x+3=1⇔x=-2`(thỏa mãn)
`->x+3=-3⇔x=-6`(thỏa mãn)
`->x+3=3⇔x=0`(thỏa mãn)
Vậy `x∈{0;-2;-4;-6}` thì `B∈Z`
