a,\(\dfrac{3x^2+3x}{x^3-x}\) = \(\dfrac{3x\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)}\) = \(\dfrac{3}{x-1}\)
b, f(2) = \(\dfrac{3}{-2-1}\) = 3 : -3 = -1
c, để A nguyên thì 3 ⋮ (x - 1)
=> (x - 1) ∈ Ư(3) = {3 ; 1 ; -1 ; -3}
x - 1 = 1 => x = 2
x - 1 = 3 => x = 4
x - 1 = -1 => x = 0
x - 1 = -3 => x = -2
=> x ∈ {2 ; -2 ; 4 ; 0}