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# Rationalise the denominator of each of the following.

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) $\frac{4}{2+\sqrt{3}+\sqrt{7}}$

Solution:

(i) $\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}}$

$=\frac{1}{(\sqrt{7}+\sqrt{6})-\sqrt{13}} \times \frac{(\sqrt{7}+\sqrt{6})+\sqrt{13}}{(\sqrt{7}+\sqrt{6})+\sqrt{13}}$

$=\frac{(\sqrt{7}+\sqrt{6})+\sqrt{13}}{(\sqrt{7}+\sqrt{6})^{2}-(\sqrt{13})^{2}}$

$=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{(\sqrt{7})^{2}+(\sqrt{6})^{2}+2(\sqrt{7})(\sqrt{6})-(\sqrt{13})^{2}}$

$=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{7+6+2 \sqrt{42}-13}$

$=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2 \sqrt{42}}$

$=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2 \sqrt{42}} \times \frac{\sqrt{42}}{\sqrt{42}}$

$=\frac{7 \sqrt{6}+6 \sqrt{7}+(\sqrt{13})(\sqrt{42})}{84}$

$=\frac{7 \sqrt{6}+6 \sqrt{7}+\sqrt{546}}{84}$

Hence, the rationalised form is $\frac{7 \sqrt{6}+6 \sqrt{7}+\sqrt{546}}{84}$.

(ii) $\frac{3}{\sqrt{3}+\sqrt{5}-\sqrt{2}}$

$=\frac{3}{(\sqrt{3}-\sqrt{2})+\sqrt{5}} \times \frac{(\sqrt{3}-\sqrt{2})-\sqrt{5}}{(\sqrt{3}-\sqrt{2})-\sqrt{5}}$

$=\frac{3\{(\sqrt{3}-\sqrt{2})-\sqrt{5}\}}{(\sqrt{3}-\sqrt{2})^{2}-(\sqrt{5})^{2}}$

$=\frac{3[(\sqrt{3}-\sqrt{2})-\sqrt{5}]}{(\sqrt{3})^{2}+(\sqrt{2})^{2}-2(\sqrt{3})(\sqrt{2})-(\sqrt{5})^{2}}$

$=\frac{3[(\sqrt{3}-\sqrt{2})-\sqrt{5}]}{3+2-2 \sqrt{6}-5}$

$=\frac{3[(\sqrt{3}-\sqrt{2})-\sqrt{5}]}{-2 \sqrt{6}}$

$=\frac{3[(\sqrt{3}-\sqrt{2})-\sqrt{5}]}{-2 \sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}$

$=\frac{3[3 \sqrt{2}-2 \sqrt{3}-\sqrt{30}]}{-12}$

$=\frac{\sqrt{30}+2 \sqrt{3}-3 \sqrt{2}}{4}$

Hence, the rationalised form is $\frac{\sqrt{30}+2 \sqrt{3}-3 \sqrt{2}}{4}$.

(iii) $\frac{4}{2+\sqrt{3}+\sqrt{7}}$

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

$=\frac{4[(2+\sqrt{3})-\sqrt{7}]}{(2+\sqrt{3})^{2}-(\sqrt{7})^{2}}$

$=\frac{4[(2+\sqrt{3})-\sqrt{7}]}{(2)^{2}+(\sqrt{3})^{2}+2(2)(\sqrt{3})-(\sqrt{7})^{2}}$

$=\frac{4[(2+\sqrt{3})-\sqrt{7}]}{4+3+4 \sqrt{3}-7}$

$=\frac{4[(2+\sqrt{3})-\sqrt{7}]}{4 \sqrt{3}}$

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

$=\frac{(2+\sqrt{3})-\sqrt{7}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$=\frac{2 \sqrt{3}+3-\sqrt{21}}{3}$

Hence, the rationalised form is $\frac{2 \sqrt{3}+3-\sqrt{21}}{3}$.

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