The image contains problems related to square roots and simplifications. I will work through the following two sections in detail:

13. Να κάνετε τις πράξεις:

This requires simplifying the following expressions.

(α)
$\sqrt{2} (\sqrt{8} + \sqrt{18})$

• Simplify each square root:
  $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}, \quad \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$
• Substitute back into the expression:
  $\sqrt{2}(2\sqrt{2} + 3\sqrt{2}) = \sqrt{2} \times 5\sqrt{2}$
• Apply multiplication:
  $= 5 \times \sqrt{2 \times 2} = 5 \times 2 = 10$

(β)
$\sqrt{8} - 3 \sqrt{200} - \sqrt{50}$

• Simplify each square root:
  $\sqrt{8} = 2\sqrt{2}, \quad \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2}, \quad \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$
• Substitute back into the expression:
  $2\sqrt{2} - 3 \times 10\sqrt{2} - 5\sqrt{2} = 2\sqrt{2} - 30\sqrt{2} - 5\sqrt{2}$
• Combine like terms:
  $(2 - 30 - 5)\sqrt{2} = -33\sqrt{2}$

14. Να αποδείξετε ότι:

These are proof exercises.

(α)
$\sqrt{2} \cdot \sqrt{20} = 2\sqrt{10}$

• Simplify the left-hand side:
  $\sqrt{2 \times 20} = \sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$
  Thus, the left-hand side equals the right-hand side.

(β)
$\frac{\sqrt{3} \cdot \sqrt{6}}{\sqrt{75}} = 3\sqrt{2}$

• Simplify each term:
  $\sqrt{3} \cdot \sqrt{6} = \sqrt{18} = 3\sqrt{2}, \quad \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$
• Substitute into the expression:
  $\frac{3\sqrt{2}}{5\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{6}}{15} = \frac{\sqrt{6}}{5}$

Would you like to explore these or other parts of the image in more detail? Here are a few related questions for practice:

1. Can you simplify $\sqrt{12} + \sqrt{27}$?
2. What is the value of $\sqrt{18} - \sqrt{8}$?
3. Show that $\sqrt{48} = 4\sqrt{3}$.
4. Simplify $\frac{\sqrt{50}}{\sqrt{2}}$.
5. Prove that $\sqrt{5} \cdot \sqrt{20} = 10$.

Tip: When simplifying square roots, look for perfect squares as factors, which makes the process faster.