Question

Please answer both the questions Please....

Answer 1

your question's answer

Answer 2

Hey friend,
Here is the answer you were looking for:

Identities used :

$(x + y)(x - y) = {x}^{2} - {y}^{2} \\ \\ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \\ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy$

$(1) \\ a = \frac{ \sqrt{3} + \sqrt{1} }{ \sqrt{3} - \sqrt{1} }$


On rationalizing the denominator we get :

$a = \frac{ \sqrt{3} + \sqrt{1} }{ \sqrt{3} - \sqrt{1} } \times \frac{ \sqrt{3} + \sqrt{1} }{ \sqrt{3} + \sqrt{1} } \\ \\ a = \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{1}) }^{2} + 2( \sqrt{3})( \sqrt{1} ) }{ {( \sqrt{3} })^{2} - {( \sqrt{1} )}^{2} } \\ \\ a = \frac{3 + 1 + 2 \sqrt{3} }{3 - 2} \\ \\ a = \frac{4 + 2 \sqrt{3} }{2} \\ \\ a = 2 + \sqrt{3}$

$b = \frac{ \sqrt{3} - \sqrt{1} }{ \sqrt{3} + \sqrt{1} }$

On rationalizing the denominator we get :

$b = \frac{ \sqrt{3} - \sqrt{1} }{ \sqrt{3} + \sqrt{1} } \times \frac{ \sqrt{3} - \sqrt{1} }{ \sqrt{3} - \sqrt{1} } \\ \\ b = \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{1} )}^{2} - 2( \sqrt{3})( \sqrt{1} ) }{ {( \sqrt{3}) }^{2} - {( \sqrt{1}) }^{2} } \\ \\ b = \frac{3 + 1 - 2 \sqrt{3} }{3 - 1} \\ \\ b = \frac{4 - 2 \sqrt{3} }{2} \\ \\ b = 2 - \sqrt{3}$

Now, putting the values :

${a}^{2} + ab - {b}^{2}$

${(2 + \sqrt{3}) }^{2} + (2 + \sqrt{3} )(2 - \sqrt{3} ) - {(2 - \sqrt{3}) }^{2} \\ \\ = ( {(2)}^{2} + {( \sqrt{3}) }^{2} + 2(2)( \sqrt{3} )) +( {(2)}^{2} - {( \sqrt{3}) }^{2} ) - ( {(2)}^{2} + {( \sqrt{3} )}^{2} - 2(2)( \sqrt{3} ) \\ \\ = (4 + 3 + 4 \sqrt{3} ) + (4 - 3) - (4 + 3 - 4 \sqrt{3} ) \\ \\ = (7 + 4 \sqrt{3} ) + (1) - (7 - 4 \sqrt{3} ) \\ \\ = 7 + 4 \sqrt{3} + 1 - 7 + 4 \sqrt{3} \\ \\ = 1 + 8 \sqrt{3}$
$(2) \\ x = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

On rationalizing the denominator we get,

$x = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ x = \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{2}) }^{2} - 2( \sqrt{3})( \sqrt{2} )}{ {( \sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} } \\ \\ x = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ \\ x = 5 - 2 \sqrt{6}$

$y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

On rationalizing the denominator we get :

$y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ y = \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{2} )}^{2} + 2( \sqrt{3} )( \sqrt{2} ) }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} } \\ \\ y = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ \\ y = 5 - 2 \sqrt{6}$


Now putting the values :

${x}^{2} + {y}^{2} + xy \\ \\ = {(5 + 2 \sqrt{6} )}^{2} + {(5 - 2 \sqrt{6}) }^{2} + (5 + 2 \sqrt{6} )(5 - 2 \sqrt{6} ) \\ \\ = ( {(5)}^{2} + {(2 \sqrt{6} )}^{2} + 2(5)(2 \sqrt{6} )) + ( {(5)}^{2} + {(2 \sqrt{6} )}^{2} - 2(5)(2 \sqrt{6} )) + ({(5)}^{2} - {(2 \sqrt{6}) }^{2} ) \\ \\ = (25 + 24 + 20 \sqrt{6} ) + (25 + 24 - 20 \sqrt{6} ) + (25 - 24) \\ \\ = 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} + (1) \\ \\ = 98 + 1 \\ \\ 99$


Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
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