Giúp mik vsssssssssss

`bt6`

`4`

`a)A=(x/(x^2-4)+1/(x+2)-2/(x-2)):(1-x/(x+2))`   `ĐKXĐ: x ne +-2 `

`A=(x/((x-2)(x+2))+(x-2)/((x-2)(x+2))-(2(x+2))/((x-2)(x+2))):(x+2-x)/(x+2)`

`A=(x+x-2-2x-4)/((x-2)(x+2)) * (x+2)/2`

`A=(-6)/(x-2)*1/2`

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

`A=3/(2-x)`

`b)` Thay `x=-4` vào biểu thức `A` ta được 

`A=3/(2-(-4))=3/(2+4)=3/6=1/2`

`c)` Để `A in ZZ` thì `3/(2-x) in ZZ`

`=>3 vdots 2-x`

`=>2-x in Ư_((3))={-3;-1;1;3}`

`=>x in {5;3;1;-1} ( tm)`

Vậy `....`