A) \[\frac{30}{31}\]
B) \[\frac{70}{31}\]
C) \[\frac{35}{31}\]
D) \[\frac{31}{37}\]
Correct Answer: D
Solution :
(d) \[y=\frac{1}{x}=\frac{1}{3+2\sqrt{2}}=\frac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}\] Expression, \[\frac{{{x}^{2}}-3xy+{{y}^{2}}}{{{x}^{2}}+3xy+{{y}^{2}}}=\frac{{{\left( x-y \right)}^{2}}-xy}{{{\left( x+y \right)}^{2}}+xy}\] \[\Rightarrow x-y=(3+2\sqrt{2})-(3-2\sqrt{2})=4\sqrt{2}\] Also, \[x+y=6\] Expression= \[\frac{32-1}{36+1}=\frac{31}{37}.\]You need to login to perform this action.
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