Question

rationalise the denominator of 1 by bracket root 7 + root 6 minus root 13

Answer 1

$\frac{1}{( \sqrt{7} + \sqrt{6} )- \sqrt{13} } = \frac{1}{ (\sqrt{7} + \sqrt{6} ) - \sqrt{13} } \times \frac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ (\sqrt{7} + \sqrt{6} ) + \sqrt{13} } \\ = \frac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ (\sqrt{7} + \sqrt{6} ) {}^{2} - ( \sqrt{13}) {}^{2} } \\ = \frac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{7 + 6 + 2 \sqrt{42} - 13 } = \frac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{2 \sqrt{42} } \\ = \frac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{2 \sqrt{42} } \times \frac{ \sqrt{42} }{ \sqrt{42} } \\ = \frac{ \sqrt{42} ( \sqrt{7} + \sqrt{6} + \sqrt{13} )}{2 \sqrt{42} \times \sqrt{42} } \\ = \frac{7 \sqrt{6} + 6 \sqrt{7} + \sqrt{546} }{2 \times 42 } \\ = \frac{7 \sqrt{6} + 6 \sqrt{7} + \sqrt{546} }{84}$