`a)`

`P = ((15-x)/(x^2 -25) + 2/(x+5)) : (x+1)/(2x^2 -10x) (ĐKXĐ: x\ne {-1;0;+-5})`

`= ((15-x)/((x-5)(x+5)) + 2/(x+5)) : (x+1)/(2x(x-5))`

`= (15-x+2(x-5))/((x-5)(x+5)) : (x+1)/(2x(x-5))`

`=(15-x+2x-10)/((x-5)(x+5)) : (x+1)/(2x(x-5))`

`=(5+x)/((x-5)(x+5)) . (2x(x-5))/(x+1)`

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

`b)`

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

Để `P \in ZZ => -2 \vdots x+1`

`=> x+1 \inƯ(-2)={+-1;+-2}`

`=> x\in{0;-2;1;-3}`

Mà `x\ne {-1;0;+-5}`

`=> x\in{-3;-2;1}`.