`a)`

`P=((2x-1)/(x+3)-x/(3-x)-(3-10x)/(x^2-9))÷(x+2)/(x-3)`

`P=[(2x-1)/(x+3)-x/(3-x)-(3-10x)/((x-3)(x+3))]·(x-3)/(x+2)`

ĐKXĐ:

`{(x-3\ne0),(x+3\ne0),(x+2\ne0):}`

`{(x\ne3),(x\ne-3),(x\ne-2):}`

`P=[(2x-1)/(x+3)-x/(3-x)-(3-10x)/((x-3)(x+3))]·(x-3)/(x+2)`

`P=[(2x-1)/(x+3)+x/(x-3)-(3-10x)/((x-3)(x+3))]·(x-3)/(x+2)`

`P=((2x-1)(x-3)+x(x+3)-3+10x)/((x-3)(x+3))·(x-3)/(x+2)`

`P=(2x^2-6x-x+3+x^2+3x-3+10x)/((x-3)(x+3))·(x-3)/(x+2)`

`P=(3x^2+6x)/((x-3)(x+3))·(x-3)/(x+2)`

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

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

`b)`

`x^-7x+12=0`

`x^2-3x-4x+12=0`

`x(x-3)-4(x-3)=0`

`(x-4)(x-3)=0`

`x=4 (TM)` hoặc `x=3 (KTM)`

Thay `x=4`, ta có:

`P=(3·4)/(4+3)=12/7`

`c)`

Ta có: `(3x)/(x+3)=(3x+9-9)/(x+3)=(3(x+3)-9)/(x+3)=3-9/(x+3)`

Để `P in ZZ` thì `9/(x+3) in ZZ`

`=>x+3 in Ư(9)={-1;-3;-9;1;3;9}`

`=>x in {-4;-6;-12;-2;0;3}`

Kết hợp với ĐKXĐ:

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