Ответ:

Решение:

  1. а)
    \( \frac{2}{\sqrt{7}} = \frac{2}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{2\sqrt{7}}{7} \)
  2. б)
    \( \frac{\sqrt{2}}{\sqrt{2}+1} = \frac{\sqrt{2}}{\sqrt{2}+1} \cdot \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}(\sqrt{2}-1)}{(\sqrt{2})^2 - 1^2} = \frac{2-\sqrt{2}}{2-1} = 2-\sqrt{2} \)

Ответ: а) \(\frac{2\sqrt{7}}{7}\); б) \(2-\sqrt{2}\).