Which numbers satisfy the following inequality:

$$
-1 \leq \frac{x^{2}+3 x-1}{4-x^{2}}<1
$$
I. Solution: $|x| \neq 2$, because otherwise the fractional expression would be undefined. To eliminate the fraction, we must consider the sign of the denominator.

I. Consider those $x$ for which $4-x^{2}>0$, that is, $|x|<2$. Dividing by this, $x-1<\frac{x^{2}+3 x-1}{4-x^{2}}$. This can be rewritten as:

$$
4-x^{2} < x^{2}+3 x-1
$$

From this, we get $x > -\frac{1}{2}$. However, we must also consider the condition $|x|<2$, so $-\frac{1}{2} < x < 2$.

II. Consider those $x$ for which $4-x^{2}<0$, that is, $|x|>2$. When removing the fraction, the inequalities reverse, so for $x$ we need:

$$
-4+x^{2} \geqq x^{2}+3 x-1>4-x^{2}
$$

From this, similarly to the previous case, we get $x \leqq -1$; this is not possible for any $x>2$. On the other hand, in product form, $(2 x+5)(x-1)>0$: here the second factor is negative for all remaining $x$, so dividing by it, we need $2 x+5<0$, which gives $x < -\frac{5}{2}$.

Combining the results, we have:

$$
\frac{x^{2}+3 x-1}{4-x^{2}}-1=\frac{2 x^{2}+3 x-5}{4-x^{2}}=\frac{(2 x+5)(x-1)}{(2-x)(2+x)}
$$

The factors in the first fraction change sign at $-2, -1$, and 2. Before -2, two factors are negative, between -2 and -1, one is negative, between -1 and 2, all are positive, and above 2, one is negative. We must also exclude the zeros of the denominator, so the first inequality holds for $x<-2.5$ or $-1 \leqq x<1$.

Each of the two inequalities, and thus (1), holds if $x<-2.5$ or $-1 \leqq x<1$.