Not so difficult

$\dfrac{4k+12}{k+3}=\dfrac{(k+3)\times 4}{k+3}$ If $k+3\neq 0$ , then we can divide by $k+3$ , so we get that the answer is $4$ . But if $k+3=0$ , then the fraction is not determined, because we can't divide by 0.

So there is exactly one counterexample, when $k=0-3=-3$ .