Question

1÷√3+√2-√8 rationalise
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Answer 1

Given,

${ \boxed{ \large{ \boxed{\frac{1}{ \sqrt{3} + \sqrt{2} - \sqrt{8} } }}}}$

Rationalising the term,

$= > {\boxed{\large{\frac{1}{ (\sqrt{3} + \sqrt{2}) - \sqrt{8}} }}}\\ \\ = > {\boxed{\large{\frac{( \sqrt{3} + \sqrt{2} ) + \sqrt{8} }{ {( \sqrt{3} + \sqrt{2} )}^{2} - {( \sqrt{8} )}^{2} } }}}\\ \\ = > {\boxed{\large{\frac{ \sqrt{3} + \sqrt{2} + \sqrt{8} }{(5 + 2 \sqrt{6} ) - 8} = \frac{ \sqrt{3} + \sqrt{2} + \sqrt{8} }{2 \sqrt{6} - 3 } }}} \\ \\ = > {\boxed{\large{\frac{ \sqrt{3} + \sqrt{2} + \sqrt{8} (2 \sqrt{6} + 3) }{ {(2 \sqrt{6} )}^{2} - {(3)}^{2} }}}} \\ \\ = > {\boxed{\large{\frac{2 \sqrt{18} + 2 \sqrt{12} + 2 \sqrt{48} + 3 \sqrt{3} + 3 \sqrt{2} + 3 \sqrt{8} }{24 - 9} }}} \\ \\ = > {\boxed{\large{ \frac{6 \sqrt{2} + 4 \sqrt{3} + 8 \sqrt{3} + 3 \sqrt{3} + 3 \sqrt{2} + 6 \sqrt{2} }{15}}}} \\ \\ = > {\boxed{\large{ \frac{15 \sqrt{2} + 15\sqrt{3} }{15} }}}\\ \\ = > {\boxed{\large{ \frac{15( \sqrt{2} + \sqrt{3}) }{15} = { \boxed{ \red{ \boxed{\sqrt{2} + \sqrt{3}}}}} }}}$

Hence,

${ \red{ \huge{ \text{Rationalised \:form \:of }}}} \\ { \boxed{\frac{1}{ \sqrt{3} + \sqrt{2} - \sqrt{8}}} } \: \: \: \: { \red{is}} \: \: \: \: { \large{ \boxed{\sqrt{2} + \sqrt{3}}}}$