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