giải hộ tôi bài toán này với

`a)`

`A = (2x)/(x+3) + (x+1)/(x-3) + (3-11x)/(9-x^2) (đk : x \ne 3 ; -3)`

`A = (2x(x-3))/((x+3)(x-3)) + ((x+1)(x+3))/((x-3)(x+3)) - (3 - 11x)/((x-3)(x+3))`

`A = (2x^2 - 6x + x^2 + 4x + 3 - 3 + 11x)/((x-3)(x+3))`

`A = (3x^2 + 9x)/((x-3)(x+3))`

`A = (3x(x + 3))/((x-3)(x+3))`

`A = (3x)/(x-3)`

`b)`

Để `A \in ZZ`

`=> 3x \vdots x - 3`

`=> 3x - 9 + 9 \vdots x-3`

`=> 3(x-3) + 9 \vdots x - 3`

Mà `3(x-3) \vdots x - 3`

`=> 9 \vdots x - 3`

`=> x-3 \in {+-1;+-3;+-9}`

`=> x \in {4;6;12;2;0;-6}`

Vậy `x \in {4;6;12;2;0;-6}`