3、Function $y=\frac{\sqrt{1-x^{2}}}{2+x}$ has the range of

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3、Function $y=\frac{\sqrt{1-x^{2}}}{2+x}$ has the range of
Answer: $\left[0, \frac{\sqrt{3}}{3}\right]$.
Solution: Since $x \in[-1,1], y \geq 0$; and according to $y^{2}=\frac{1-x^{2}}{(x+2)^{2}}$, let $t=x+2$, then $y^{2}=\frac{1-(t-2)^{2}}{t^{2}}$, i.e., $\left(y^{2}+1\right) t^{2}-4 t+3=0$, considering it as a quadratic equation in $t$, its discriminant $\Delta=16-12\left(y^{2}+1\right) \geq 0$, we get $y^{2} \leq \frac{1}{3}$, $|y| \leq \frac{\sqrt{3}}{3}$, since $y \geq 0$, so $y \in\left[0, \frac{\sqrt{3}}{3}\right]$.

Alternative solution: $x \in[-1,1]$, let $x=\cos \alpha, \alpha \in[0, \pi]$, then $y=\frac{\sin \alpha}{2+\cos \alpha}, y \geq 0$, $2 y=\sin \alpha-y \cos \alpha=\sqrt{1+y^{2}} \cdot \sin (\alpha+\theta) \leq \sqrt{1+y^{2}}$, so $1+y^{2} \geq 4 y^{2}$, i.e., $y^{2} \leq \frac{1}{3}$, since $y \geq 0$, then $y \in\left[0, \frac{\sqrt{3}}{3}\right]$.