The value of $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is:

Let's simplify the radical terms in both the numerator and the denominator:
1. Numerator:

$$\sqrt{32}=\sqrt{16\times 2}=4\sqrt{2}$$

$$\sqrt{48}=\sqrt{16\times 3}=4\sqrt{3}$$

$$\text{Numerator}=4\sqrt{2}+4\sqrt{3}=4(\sqrt{2}+\sqrt{3})$$

2. Denominator:

$$\sqrt{8}=\sqrt{4\times 2}=2\sqrt{2}$$

$$\sqrt{12}=\sqrt{4\times 3}=2\sqrt{3}$$

$$\text{Denominator}=2\sqrt{2}+2\sqrt{3}=2(\sqrt{2}+\sqrt{3})$$

3. Divide Numerator by Denominator:

$$\frac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})}=\frac{4}{2}=2$$