Question

Rationalise the denominator of 1/√3+√2​

Answer 1

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✴ANSWER ⤵⤵⤵⤵

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

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Answer 2

Answer:

Question :

$1 \div \sqrt{3} + \sqrt{2}$

Step-by-step explanation:

for rationalisation we have to multiply and divide by the denominator.

Note: the sign between the terms must be reversed while multiplying and dividing.

So according to the rationalisation rule we have:

$(1 \div \sqrt{3} + \sqrt{2} ) \times (( \sqrt{3} - \sqrt{2} ) \div \sqrt{3} - \sqrt{2} ))$

$= >( (\sqrt{3} - \sqrt{2} ) \div { \sqrt{3} }^{2} - { \sqrt{2} }^{2} ))$

$( (\sqrt{3} - \sqrt{2} ) \div( 3 - 2))$

$= > \sqrt{3} - \sqrt{2}$

which is the required answer

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