Question

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

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Solution

## $\left(\mathrm{i}\right)\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\sqrt{7}+\sqrt{6}\right)-\sqrt{13}}×\frac{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}\phantom{\rule{0ex}{0ex}}=\frac{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}{{\left(\sqrt{7}+\sqrt{6}\right)}^{2}-{\left(\sqrt{13}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{{\left(\sqrt{7}\right)}^{2}+{\left(\sqrt{6}\right)}^{2}+2\left(\sqrt{7}\right)\left(\sqrt{6}\right)-{\left(\sqrt{13}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{7+6+2\sqrt{42}-13}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}×\frac{\sqrt{42}}{\sqrt{42}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{6}+6\sqrt{7}+\left(\sqrt{13}\right)\left(\sqrt{42}\right)}{84}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}$ Hence, the rationalised form is $\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}$. $\left(\mathrm{ii}\right)\frac{3}{\sqrt{3}+\sqrt{5}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{3}{\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{5}}×\frac{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}}{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{3\left\{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right\}}{{\left(\sqrt{3}-\sqrt{2}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{{\left(\sqrt{3}\right)}^{2}+{\left(\sqrt{2}\right)}^{2}-2\left(\sqrt{3}\right)\left(\sqrt{2}\right)-{\left(\sqrt{5}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{3+2-2\sqrt{6}-5}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{-2\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{-2\sqrt{6}}×\frac{\sqrt{6}}{\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[3\sqrt{2}-2\sqrt{3}-\sqrt{30}\right]}{-12}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{30}+2\sqrt{3}-3\sqrt{2}}{4}$ Hence, the rationalised form is $\frac{\sqrt{30}+2\sqrt{3}-3\sqrt{2}}{4}$. $\left(\mathrm{iii}\right)\frac{4}{2+\sqrt{3}+\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{4}{\left(2+\sqrt{3}\right)+\sqrt{7}}×\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\left(2+\sqrt{3}\right)-\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{{\left(2+\sqrt{3}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{{\left(2\right)}^{2}+{\left(\sqrt{3}\right)}^{2}+2\left(2\right)\left(\sqrt{3}\right)-{\left(\sqrt{7}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{4+3+4\sqrt{3}-7}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}+3-\sqrt{21}}{3}$ Hence, the rationalised form is $\frac{2\sqrt{3}+3-\sqrt{21}}{3}$.

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