Question

Number system--------

Answer 1

Here is your answer
See down
In pic




Hope it helped you
If you like my answer then please mark it has brainliest answer and rate it

Answer 2

Hey friend,
Here is the answer you were looking for:
$x = \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1 }$

On rationalizing the denominator we get,

$x = \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1} \times \frac{ \sqrt{2} + 1}{ \sqrt{2} + 1 }$

Using the identities:
${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ (x + y)(x - y) = {x}^{2} - {y}^{2}$


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

$y = \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1}$

On rationalizing the denominator we get,

$y = \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1 } \times \frac{ \sqrt{2} - 1 }{ \sqrt{2} - 1}$

Using the identities:

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

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

${x}^{2} + {y}^{2} + xy$

Now,
putting the values we get:

${(3 + 2 \sqrt{2}) }^{2} + {(3 - 2 \sqrt{2} )}^{2} + (3 + 2 \sqrt{2} )(3 - 2 \sqrt{2} )$

Using same identities as above:

$( {(3)}^{2} + {(2 \sqrt{2}) }^{2} + 2(3)(2 \sqrt{2} )) + ( {(3)}^{2} + {(2 \sqrt{2}) }^{2} - 2(3)(2 \sqrt{2} ) + ( {(3)}^{2} - {(2 \sqrt{2} )}^{2} ) \\ \\ = (9 + 8 + 12 \sqrt{2} ) + (9 + 8 - 12 \sqrt{2} ) + (9 - 8) \\ \\ = 17 + 12 \sqrt{2} + 17 - 12 \sqrt{2} + 1 \\ \\ = 34 + 1 \\ \\ = 35$

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...
☺☺