Correct option is D

Given:

​$$x = \frac{\sqrt{5} - \sqrt{4}}{\sqrt{5} + \sqrt{4}}, \quad y = \frac{\sqrt{5} + \sqrt{4}}{\sqrt{5} - \sqrt{4}}$$​​

We are required to find the value of: $$\frac{x^2 - xy + y^2}{x^2 + xy + y^2}$$​​

Concept Used:

First, note that x × y = 1 

$$a + \frac{1}{a} = k \\\ \\a^2 + \frac{1}{a^2} = k^2 - 2$$​​​

Substitute xy = 1 in the above expression.

Solution:

Given

x × y = 1  

​$$y = \frac{1}{x}$$ 

$$x = \frac{\sqrt{5} - \sqrt{4}}{\sqrt{5} + \sqrt{4}},$$ 

$$x = \frac{\sqrt{5} - \sqrt{4}}{\sqrt{5} + \sqrt{4}}\times\frac{\sqrt{5} - \sqrt{4}}{\sqrt{5} - \sqrt{4}}, \\\ \\ x = \frac{(\sqrt{5} - \sqrt{4})^2}{(\sqrt{5})^2 - (\sqrt{4})^2}\\\ \\x = 5 + 4 - 2\sqrt{20} \\\ \\x = 9 - 4\sqrt{5}$$

$$\frac{1}{x} = \frac{1}{ 9 - 4\sqrt5} \times \frac{9 + 4\sqrt5}{ 9 + 4\sqrt5} \\\ \\\frac{1}{x} = \frac{9 +4\sqrt5}{81 - 80} \\\ \\\frac{1}{x} = 9 + 4\sqrt{5}$$​   

Then the value of $$x + \frac{1}{x} = 9 - 4\sqrt{5} + 9 + 4\sqrt{5}$$ 

$$x +\frac{1}{x} = 18$$ 

$$x^2 + \frac{1}{x^2 } = {18}^2 - 2 \\\ \\x^2 + \frac{1}{x^2 } = 324 - 2 \\\ \\x^2 + \frac{1}{x^2 } = 322$$

​$$= \frac{x^2 + \frac{1}{x^2} - 1}{x^2 + \frac{1}{x^2} + 1}$$  

= $$\frac{322 - 1}{322 +1}$$ 

= $$\frac{321}{323}$$​​

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