Đáp án:
`1)P=3/(x-3)-x/(x-3)+(x^2+9)/(x^2-9)(x ne +-3)`
`<=>P=(3-x)/(x-3)+(x^2+9)/(x^2-9)`
`<=>P=-1+(x^2+9)/(x^2-9)`
Để `P in ZZ`
`=>x^2+9 \vdots x^2-9`
`=>x^2-9+18 vdots x^2-9`
`=>18 vdots x^2-9`
`=>x^2-9 in Ư(18)={+-1,+-2,+-3,+-6,+-9,+-18}`
Mà `x^2-9>=-9`
`=>x^2-9 in {+-1,+-2,+-3,+-6,+-9}`
`=>x^2 in {10,8,7,11,6,12,3,15,0,18}`
Mà `x in ZZ`
`=>x=0`.
Vậy `x=0` thì `P in ZZ`.