Question

Rationalise the denominator of each of the following.
(i) $\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}}$ (ii) $\frac{3}{\sqrt{3}{\displaystyle +}{\displaystyle \sqrt{5}}{\displaystyle -}{\displaystyle \sqrt{2}}}$ (iii) $\frac{4}{2+\sqrt{3}+\sqrt{7}}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}} \\ =\frac{1}{\left(\sqrt{7}+\sqrt{6}\right)-\sqrt{13}}\times \frac{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}} \\ =\frac{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}{{\left(\sqrt{7}+\sqrt{6}\right)}^2-{\left(\sqrt{13}\right)}^2} \\ =\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{{\left(\sqrt{7}\right)}^2+{\left(\sqrt{6}\right)}^2+2\left(\sqrt{7}\right)\left(\sqrt{6}\right)-{\left(\sqrt{13}\right)}^2} \\ =\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{7+6+2\sqrt{42}-13} \\ =\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}} \\ =\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}\times \frac{\sqrt{42}}{\sqrt{42}} \\ =\frac{7\sqrt{6}+6\sqrt{7}+\left(\sqrt{13}\right)\left(\sqrt{42}\right)}{84} \\ =\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84} \end{gathered}$$
Hence, the rationalised form is $\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}$.
$$\begin{gathered} \left(\mathrm{ii}\right)\frac{3}{\sqrt{3}+\sqrt{5}-\sqrt{2}} \\ =\frac{3}{\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{5}}\times \frac{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}}{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}} \\ =\frac{3\left\{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right\}}{{\left(\sqrt{3}-\sqrt{2}\right)}^2-{\left(\sqrt{5}\right)}^2} \\ =\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2-2\left(\sqrt{3}\right)\left(\sqrt{2}\right)-{\left(\sqrt{5}\right)}^2} \\ =\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{3+2-2\sqrt{6}-5} \\ =\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{-2\sqrt{6}} \\ =\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{-2\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}} \\ =\frac{3\left[3\sqrt{2}-2\sqrt{3}-\sqrt{30}\right]}{-12} \\ =\frac{\sqrt{30}+2\sqrt{3}-3\sqrt{2}}{4} \end{gathered}$$
Hence, the rationalised form is $\frac{\sqrt{30}+2\sqrt{3}-3\sqrt{2}}{4}$.
$$\begin{gathered} \left(\mathrm{iii}\right)\frac{4}{2+\sqrt{3}+\sqrt{7}} \\ =\frac{4}{\left(2+\sqrt{3}\right)+\sqrt{7}}\times \frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\left(2+\sqrt{3}\right)-\sqrt{7}} \\ =\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{{\left(2+\sqrt{3}\right)}^2-{\left(\sqrt{7}\right)}^2} \\ =\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{{\left(2\right)}^2+{\left(\sqrt{3}\right)}^2+2\left(2\right)\left(\sqrt{3}\right)-{\left(\sqrt{7}\right)}^2} \\ =\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{4+3+4\sqrt{3}-7} \\ =\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{4\sqrt{3}} \\ =\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\sqrt{3}} \\ =\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ =\frac{2\sqrt{3}+3-\sqrt{21}}{3} \end{gathered}$$
Hence, the rationalised form is $\frac{2\sqrt{3}+3-\sqrt{21}}{3}$.