Question

Rationalize the denominator of $\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}}.$

Answer 1

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

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

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

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

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

$=\frac{\left(\sqrt{7}+\sqrt{6}+\sqrt{13}\right)\sqrt{42}}{2\times 42}$

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