Question

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Answer 1

Answer :

The required value is 63/61

Step-by-step explanation :

To solve this question, first we have to find the values of x and y by rationalizing their denominators.

• The rationalizing factor of x is (√5 - √3)
• The rationalizing factor of y is (√5 + √3)

Rationalizing the denominator of x :

    $\sf x=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\times \dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}\\\\ \sf x=\dfrac{(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}\\\\ \sf x=\dfrac{\sqrt{5}(\sqrt{5}-\sqrt{3})-\sqrt{3}(\sqrt{5}-\sqrt{3})}{\sqrt{5}(\sqrt{5}-\sqrt{3})+\sqrt{3}(\sqrt{5}-\sqrt{3})}\\\\ \sf x=\dfrac{\sqrt{5}^2-\sqrt{15}-\sqrt{15}+\sqrt{3}^2}{\sqrt{5}^2-\sqrt{15}+\sqrt{15}-\sqrt{3}^2}\\\\ \sf x=\dfrac{5-2\sqrt{15}+3}{5-3}\\\\ \sf x=\dfrac{8-2\sqrt{15}}{2}$

    $\boxed{\sf x=4-\sqrt{15}}$

Rationalizing the denominator of y :

  $\sf y=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\times \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}\\\\ \sf y=\dfrac{(\sqrt{5}+\sqrt{3})(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}\\\\ \sf y=\dfrac{\sqrt{5}(\sqrt{5}+\sqrt{3})+\sqrt{3}(\sqrt{5}+\sqrt{3})}{\sqrt{5}(\sqrt{5}+\sqrt{3})-\sqrt{3}(\sqrt{5}+\sqrt{3})}\\\\ \sf y=\dfrac{\sqrt{5}^2+\sqrt{15}+\sqrt{15}+\sqrt{3}^2}{\sqrt{5}^2+\sqrt{15}-\sqrt{15}-\sqrt{3}^2}\\\\ \sf y=\dfrac{5+2\sqrt{15}+3}{5-3}\\\\ \sf y=\dfrac{8+2\sqrt{15}}{2}$

  $\boxed{\sf y=4+\sqrt{15}}$

Now,

x² = (4 - √15)²

x² = 4² + (√15)² - 2(4)(√15)

x² = 16 + 15 - 8√15

x² = 31 - 8√15

 

y² = (4 + √15)²

y² = 4² + (√15)² + 2(4)(√15)

y² = 16 + 15 + 8√15

y² = 31 + 8√15

xy = (4 - √15) (4 + √15)

xy = 4² - √15²

xy = 16 - 15

xy = 1

Used identities :

(a - b)² = a² + b² - 2ab

(a + b)² = a² + b² + 2ab

(a + b) (a - b) = a² - b²

__________________

Substitute the values,

   $\sf \longrightarrow \dfrac{x^2+xy+y^2}{x^2-xy+y^2} \\\\\\ \sf \longrightarrow \dfrac{31-8\sqrt{15}+1+31+8\sqrt{15}}{31-8\sqrt{15}-1+31+8\sqrt{15}} \\\\\\ \sf \longrightarrow \dfrac{31+1+31}{31-1+31} \\\\\\ \sf \longrightarrow \dfrac{63}{61}$

Therefore, the required value is 63/61

Answer 2

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