$\textit{Đáp án + Giải thích các bước giải:}$

$a,$ `M=2/(x+3)+(-1)/(x-3)+(2x)/(x^2-9)` `(x \ne +-3)`

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

`=(2x-6-x-3+2x)/((x-3)(x+3))`

`=(3x-9)/((x-3)(x+3))`

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

`=3/(x+3)` 

$b,$ `P=M+N`

`<=>P=3/(x+3)+(2x)/(x+3)`

`<=>P=(3+2x)/(x+3)`

Để `P` nhận giá trị nguyên thì `3+2x \vdots x+3`

`<=>2x+6-3 \vdots x+3`

`<=>2(x+3)-3 \vdots x+3`

`<=>3 \vdots x+3`

`<=>x+3 \in Ư_{(3)}={-3;-1;1;3}`

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

Vậy `x \in {-6;-4;-2;0}` thì `P` nguyên

ĐKXĐ : `x\ne 3,x\ne -3`

`a,`

`M=2/(x+3)+(-1)/(x-3)+(2x)/(x^2-9)`

`=(2(x-3) - (x+3) +2x)/((x-3)(x+3))`

`=(2x-6 - x-3+2x)/((x-3)(x+3))`

`= (3x  - 9)/((x-3)(x+3))`

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

`=3/(x+3)`

Vậy `M=3/(x+3)`

`b,`

`P=M+N`

`=3/(x+3)+(2x)/(x+3)`

`=(3+2x)/(x+3)`

`=(2x+6 -3)/(x+3)`

`=(2(x+3)-3)/(x+3)`

`= 2 - 3/(x+3)`

Để `P` nguyên

`->3\vdots x+3`

`->x+3\in Ư (3)={1;-1;3;-3}`

`->x\in {-2; -4; 0; -6}` (Tm)

Vậy `x\in {-2; -4; 0;-6}` để `P` nguyên