Question

Rationalise the denominator of the following

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Answer 1

a)

$\frac{1}{3 + \sqrt{2} } = \frac{1}{3 + \sqrt{2} } \times \frac{3 - \sqrt{2} }{3 - \sqrt{2} } = \frac{3 - \sqrt{2} }{(3 + \sqrt{2})(3 - \sqrt{2} ) } = \frac{3 - \sqrt{2} }{ {(3)}^{2} - { (\sqrt{2}) }^{2} } = \frac{3 - \sqrt{2} }{9 - 2} = \frac{3 - \sqrt{2} }{7}$

b)

$\frac{1}{ \sqrt{6} - \sqrt{5} } = \frac{1}{ \sqrt{6 } - \sqrt{5} } \times \frac{ \sqrt{6} + \sqrt{5} }{ \sqrt{6} + \sqrt{5} } = \frac{ \sqrt{6} + \sqrt{5} }{( \sqrt{6} - \sqrt{5})( \sqrt{6} + \sqrt{5} ) } = \frac{ \sqrt{6} + \sqrt{5} }{( { \sqrt{6} })^{2} - ( { \sqrt{5} })^{2} } = \frac{ \sqrt{6} + \sqrt{5} }{6 - 5} = \frac{ \sqrt{6} + \sqrt{5} }{1} = \sqrt{6} + \sqrt{5}$

c)

$\frac{1}{ \sqrt{41} - 5} = \frac{1}{ \sqrt{41} - 5} \times \frac{ \sqrt{41} + 5}{ \sqrt{41} + 5} = \frac{ \sqrt{41} + 5 }{( \sqrt{41} - 5)( \sqrt{41} + 5 )} = \frac{ \sqrt{41} + 5 }{( { \sqrt{41}) }^{2} - {(5)}^{2} } = \frac{ \sqrt{41} + 5}{41 - 25} = \frac{ \sqrt{41} + 5 }{16}$