A) \[\frac{30}{31}\]
B) \[\frac{70}{31}\]
C) \[\frac{35}{31}\]
D) \[\frac{31}{37}\]
Correct Answer: D
Solution :
(d): \[x=3+2\sqrt{2}\] \[xy=1\Rightarrow y=\frac{1}{3+2\sqrt{2}}=\frac{1}{3+2\sqrt{2}}\times \frac{3-2\sqrt{2}}{3-2\sqrt{2}}\] \[=\frac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}\] \[\,\therefore x+y=3+2\sqrt{2}+3-2\sqrt{2}=6\] \[\therefore \frac{{{x}^{2}}-3xy+{{y}^{2}}}{{{x}^{2}}+3xy+{{y}^{2}}}=\frac{{{\left( x+y \right)}^{2}}-5xy}{{{\left( x+y \right)}^{2}}+xy}=\frac{36-5}{36+1}=\frac{31}{37}\]
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