3. Let's call white numbers those of the form $\sqrt{a+b \sqrt{2}}$, where $\mathbf{a}$ and $\mathbf{b}$ are integers, not equal to zero. Similarly, let's call black numbers those of the form $\sqrt{c+d \sqrt{7}}$, where $c$ and $d$ are integers, not equal to zero. Can a black number equal the sum of several white numbers?
Solution We will look for a black number equal to the sum of two white ones. Let's take conjugate white numbers:

$\sqrt{A+B \sqrt{2}}+\sqrt{A-B \sqrt{2}}=\sqrt{C+D \sqrt{7}}$

Squaring both sides of the equation, we get:

$$
2 A+2 \sqrt{A^{2}-2 B^{2}}=C+D \sqrt{7}
$$

Thus, it is sufficient to choose such numbers $A$ and $B$ that $A^{2}-2 B^{2}=7$, then we can take $C=2 A$, $D=2$. For example, if $A=3, B=1, C=6, D=2$, then we have:

$\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}=\sqrt{6+2 \sqrt{7}}$.

This equality is easily verified by squaring.

Answer: it can.