Bài `1:`

`1)`

`A=(3/(x-1)-2/(x+1)) : (x+2)/(x^2-1) \ (xne+-1;xne-2)`

`A=[(3(x+1))/((x-1)(x+1))-(2(x-1))/((x-1)(x+1))] : (x+2)/((x-1)(x+1))`

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

`A=(x+5)/((x-1)(x+1)) * ((x-1)(x+1))/(x+2)`

`A=(x+5)/(x+2)`

`2)`

Ta có: `A=(x+5)/(x+2)=(x+2+3)/(x+2)=1+3/(x+2)`

Để `A` nguyên thì `3/(x+2)` nguyên

`=> (x+2) in Ư(3)`

`=> Ư(3)={+-1;+-3}`

Ta có bảng: Hình.

Vậy `x in {-3;-5}` thì `A in ZZ`

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