Question

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$\sf \: if \: \sf\mathrm{x=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}} \: \: and \: \: \mathrm{y=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}} \\ \\ \\ \sf \: find \: the \: value \: of \: \sf{\dfrac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}}$
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Answer 1

Answer:

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$\\ \sf x = \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \: \: \: </p><p> and \: \: \sf y = \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

To find :

$\sf \: The \: value \: of \: \longrightarrow \sf\dfrac{x {}^{2} + xy + y {}^{2} }{x {}^{2} - xy + y {}^{2} }$

Formula's used:

$\sf1) \: \: \: \: (x + y)(x - y) = x {}^{2} - y {}^{2}1)(x+y)(x−y)$

$\sf \: 2) \: \: \sf(x + y) {}^{2} = x {}^{2} + y {}^{2} + 2xy2)(x+y)$

$3) \: \: \sf(x -y) {}^{2} = x {}^{2} + y {}^{2} - 2xy {3}^{} )(x−y)$

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$\sf x = \dfrac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} + \sqrt{3} }$

Now rationalise the denominator

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

We know that (a+b)(a-b)=a²-b²

$\implies \sf x = \frac{( \sqrt{5} - \sqrt{3} ) {}^{2} }{( \sqrt{5}) {}^{2} - ( \sqrt{3} ) {}^{2} }$

we know that (a-b)²= a ² +b²-2ab

$\implies \sf x = \frac{5 + 3 - 2 \sqrt{3} \times \sqrt{5} }{2}\\ \\ \sf\implies \sf x = \frac{8 - 2 \sqrt{15} }{2}\\ \\ \sf \implies \sf x = 4 - \sqrt{15} ...(1)\\ \\ \sf y = \dfrac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

Rationalise the denominator

$\implies \sf y = \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} }$

$\sf \implies y = \frac{( \sqrt{5} + \sqrt{3}) {}^{2} }{2}\\ \\ \sf\implies \sf y = \frac{5 + 3 + 2 \sqrt{15} }{2}\\ \\ \sf\implies \sf y = \frac{8 + 2 \sqrt{15} }{2}\\ \\ \sf\implies \sf y = 4 + \sqrt{15} \: \: \: \: ...(2)$

We have to find the value of

$\longrightarrow\sf \dfrac{x {}^{2} + xy + y {}^{2} }{x {}^{2} - xy + y {}^{2} }$

First solve Numerator

$\: \\ \longrightarrow\sf x {}^{2} + xy + y {}^{2}$

Put the values of x and y from equation (1) and (2)

$\sf \: = \sf (4 - \sqrt{15}) {}^{2} + (4 - \sqrt{15} )(4 + \sqrt{15} ) + (4 + \sqrt{15} ) {}^{2}$

$\sf = (16 + 15 - 2 \times 4 \times \sqrt{15} ) + (4 {}^{2} - (\sqrt{15}) {}^{2} ) + (16 + 15 + 2 \times 4 \times \sqrt{15} )$

$= 31 - \cancel{8 \sqrt{15} } - 1 + 31 + \cancel{8 \sqrt{15} }$

$\sf = 62 + 1 = 63=62+1=63$

Now solve denominator

$\longrightarrow\sf x {}^{2} - xy + y {}^{2}$

Put the values of x and y from equation (1) and (2)

$\sf = (4 - \sqrt{15}) {}^{2} - (4 {}^{2} - ( \sqrt{15}) {}^{2} ) + (4 + \sqrt{3}) {}^2$

$\sf = 16 + 15 - 8 \sqrt{15} - (16 - 15) + 16 + 15 + 8 \sqrt{15}$

$\sf = 31 - \cancel{8 \sqrt{15} } - 1 + 31 + \cancel{8 \sqrt{15}}$

$\sf = 62 - 1 = 61=62−1=61$

Therefore,

$\longrightarrow\sf \dfrac{x {}^{2} + xy + y {}^{2} }{x {}^{2} - xy + y {}^{2} } = \dfrac{63}{61}$

Answer 2

Answer:

$e\red{\boxed{\pink{\mathbb{\overbrace{\underbrace{\fcolorbox{red}{aqua}{\underline{\pink{answer \: in \: attachement : -}}}}}}}}} </p><p> : -$

Step-by-step explanation:

Hope it helpful for you