a)

$ Đkxđ : x \neq ± 3 $

Ta có :` A = ( \frac{x}{x+3} + \frac{2x}{x-3} - \frac{3x^2+12}{x^2-9} ) : \frac{3}{x-3} `

              ` = \frac{ x ( x - 3 ) + 2x ( x + 3 ) - ( 3x^2 + 12 ) }{ ( x + 3 ) ( x - 3 ) } . \frac{x-3}{3} `

              ` = \frac{x^2-3x+2x^2+ 6x -3x^2-12 }{(x + 3 ) ( x -3 )} . \frac{x-3}{3} `

              ` = \frac{3x -12}{( x + 3 ) ( x - 3 )} . \frac{x-3}{3} `

              ` = \frac{3(x-4)}{( x + 3 ) ( x - 3 )} . \frac{x-3}{3} `

              ` = \frac{x-4}{x+3}`

Vậy $ A = \dfrac{x-4}{x+3}$

b)

Ta có : $ A = \dfrac{x-4}{x+3} = \dfrac{x+3-7}{x+3} = 1 - \dfrac{7}{x+3} $

Để $A$ nhận giá trị nguyên

$ ⇒ A$ nguyên

$ ⇒ \dfrac{7}{x+3} $nguyên

` ⇒ x + 3 ∈ Ư(7) = { -7 ; -1; 1 ; 7 } ` 

` ⇒ x ∈ { -10 ; -4 ; -2 ; 4 }`

Vậy ` x ∈ { -10 ; -4 ; -2 ; 4 }` thì $A$ nguyên 

$\\a,A=(\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2+12}{x^2-9}):\dfrac{3}{x-3}$

Điều kiện xác định : $\begin{cases} x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{cases}$ $⇔$ $\begin{cases} x\ne-3\\x\ne 3\end{cases}$

Khi đó :

$A=(\dfrac{x(x-3)}{(x-3)(x+3)} +\dfrac{2x(x+3)}{(x-3)(x+3)}-\dfrac{3x^2+12}{(x-3)(x+3)}).\dfrac{x-3}{3}\\=\dfrac{x^2-3x + 2x^2+6x - 3x^2-12}{(x-3)(x+3)}.\dfrac{x-3}{3}\\=\dfrac{3x-12}{3x + 9}\\=\dfrac{x-4}{x+3}$

Vậy $A=\dfrac{x-4}{x+3}$

$b,\\A=\dfrac{x-4}{x+3}(x\ne3,x\ne-3)\\A=\dfrac{x+3-7}{x+3}\\=1-\dfrac{7}{x+3}$

Để $A$ nguyên

`⇒ 7\vdots x+3`

`⇒x+3\in Ư (7)={1;-1;7;-7}`

`⇒ x ∈ {-2;-4;4;-10}`

Vậy `x\in {-2;-4;4;-10}` để `A` nguyên