Rationalise the denominators of the following: <br/> <br/> (i)$\frac{1}{\sqrt{7}}$<br/> <br/> (ii)$\frac{1}{\sqrt{7}-\sqrt{6}}$ <br/> <br/>(iii) $\frac{1}{\sqrt{5}+\sqrt{2}}$ <br/> <br/> (iv) $\frac{1}{\sqrt{7}-2}$

Solution:

(i) $\frac{1}{\sqrt{7}}=\frac{1 \times \sqrt{7}}{1 \times \sqrt{7}}=\frac{\sqrt{7}}{7}$

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

$=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}$

$=\frac{\sqrt{7}+\sqrt{6}}{7-6}=\frac{\sqrt{7}+\sqrt{6}}{1}=\sqrt{7}+\sqrt{6}$

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

$=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}=\frac{\sqrt{5}-\sqrt{2}}{5-2}$

$=\frac{\sqrt{5}-\sqrt{2}}{3}$

(iv) $\frac{1}{\sqrt{7}-2}=\frac{1}{(\sqrt{7}-2)(\sqrt{7}+2)}$

$=\frac{\sqrt{7}+2}{(\sqrt{7})^{2}-(2)^{2}}$

$=\frac{\sqrt{7}+2}{7-4}=\frac{\sqrt{7}+2}{3}$