Question

rationalise the denominator 9/√2+√3​

Answer 1

ANSWER:

$\frac{9}{ \sqrt{2} + \sqrt{3} }$

$= \frac{9}{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} }$

$= \frac{9( \sqrt{2} - \sqrt{3} )}{ { (\sqrt{2} )}^{2} - {( \sqrt{3}) }^{2} }$

$= \frac{9( \sqrt{2} + \sqrt{3} ) }{2 - 3}$

$= \frac{9( \sqrt{2} - \sqrt{3} ) }{ - 1}$

$= - 9( \sqrt{2} - \sqrt{3} )$

$= - 9 \sqrt{2} + 9 \sqrt{3}$

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