Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$? Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$ ? 


$$\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d\in\mathbf{Q}\}$$ 

$$\mathbf{Q}(\sqrt{2}+\sqrt{3}) = \lbrace a+b(\sqrt{2}+\sqrt{3}) \mid a,b \in \mathbf{Q} \rbrace  $$

So if an element is in $\mathbf Q(\sqrt{2},\sqrt{3})$, then it is in $\mathbf{Q}(\sqrt{2}+\sqrt{3})$, because $\sqrt{6} = \sqrt{2}\sqrt{3}$.

How to conclude from there?
$\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is clear.

Now note that $$(\sqrt{2} + \sqrt{3})^{-1} = \frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{2 -  3} = \sqrt{3} - \sqrt{2}$$ hence $\sqrt{3} - \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence 
$\sqrt{2} + \sqrt{3} + \sqrt{3} - \sqrt{2} = 2 \sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Note that by a similar argument you get $\sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3}) $.
Another solution:
Clearly it has to be a subfield of $\mathbf{Q}(\sqrt{2}, \sqrt{3})$. Thus its degree has to be one of these numbers: $1,2,4$. We just have to rule out it is of degree $2$. If it is, then there is a minimal polynomial of degree two s.t $\sqrt{2}+\sqrt{3}$ is its root. However, if there was such polynomial, say $x^2+bx+c$ then $5+c+2\sqrt{6}+b(\sqrt{3}+\sqrt{2})=0$ which is impossible as $1,\sqrt{2},\sqrt{3},\sqrt{6}$ is a basis of $\mathbf{Q}(\sqrt{2}, \sqrt{3})$ and the equation implies it is linearly dependent!