Rationalize the denominator of the following :
i)\(\frac{1}{\sqrt{7}} \)
ii)\( \frac{1}{\sqrt{7} } - \sqrt{6} \)
iii)\(\frac{1}{\sqrt{5}} + \sqrt{2} \)
iv)\(\frac{1}{\sqrt{7}} - 2 \)

To rationalize the denominator, we need to multiply the irrational denominator by it's conjugate.
i) \(\frac{1}{\sqrt{7}} \) × \(\frac{\sqrt{7}}{\sqrt{7}} \)
= \(\frac{\sqrt{7}}{7} \)

ii) \(\frac{1}{(\sqrt{7} - \sqrt{6})} × \frac{(\sqrt{7} + \sqrt{6})}{(\sqrt{7} + \sqrt{6})} \)
\(= \frac{(\sqrt{7} + \sqrt{6})}{(7-6)} \)
\( = \sqrt{7} + \sqrt{6} \)

iii) \(\frac{1}{(\sqrt{5} + \sqrt{2})} \) × \(\frac{(\sqrt{5} - \sqrt{2})}{(\sqrt{5} - \sqrt{2})} \)
\( = \frac{(\sqrt{5} - \sqrt{2})}{(5 - 2)} \)
\( = \frac{(\sqrt{5} - \sqrt{2})}{3} \)

iv) \(\frac{1}{(\sqrt{7} - 2)} \) × \( \frac{(\sqrt{7} + 2)}{(\sqrt{7} + 2)} \)
\( = \frac{(\sqrt{7} + 2)}{(7 - 4)} \)
\( = \frac{\sqrt{7} + 2}{3} \)