Question

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

Answer 1

(1) $\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}$ ( multiplying and dividing by $\sqrt{7}$)

$=\frac{\sqrt{7}}{7}$

(2) $\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}\times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$ (multiplying and dividing by $\sqrt{7}+\sqrt{6}$)

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

(3) $\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{1}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ (multiplying and dividing by $\sqrt{5}-\sqrt{2}$

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

(4) $\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2}\times \frac{\sqrt{7}+2}{\sqrt{7}+2}$ (multiplying and dividing by $\sqrt{7}+2$)

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