17. Calculate $$(\sqrt{42}-2)^2 + 4\sqrt{42}$$.

Let's solve this step-by-step:

1. We need to expand the term $$(\sqrt{42}-2)^2$$ first. This is in the form $$(a-b)^2 = a^2 - 2ab + b^2$$.
2. Here, $$a = \sqrt{42}$$ and $$b = 2$$.
3. So, $$(\sqrt{42})^2 - 2(\sqrt{42})(2) + 2^2$$.
4. $$(\sqrt{42})^2 = 42$$.
5. $$2(\sqrt{42})(2) = 4\sqrt{42}$$.
6. $$2^2 = 4$$.
7. So, $$(\sqrt{42}-2)^2 = 42 - 4\sqrt{42} + 4$$.
8. Combine the constant terms: $$42 + 4 = 46$$.
9. So, $$(\sqrt{42}-2)^2 = 46 - 4\sqrt{42}$$.
10. Now, substitute this back into the original expression: $$(46 - 4\sqrt{42}) + 4\sqrt{42}$$.
11. The $$-4\sqrt{42}$$ and $$+4\sqrt{42}$$ cancel each other out.
12. This leaves us with 46.

Ответ: 46