A) \[\frac{7}{38}\]
B) \[\frac{7}{40}\]
C) \[\frac{7}{19}\]
D) \[\frac{7}{26}\]
Correct Answer: D
Solution :
\[x=2+\sqrt{3},y+2-\sqrt{3}\] \[x+y=4\] \[xy=4-3=1\] \[\therefore \] \[\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}+{{y}^{3}}}=\frac{{{(x+y)}^{2}}-2xy}{{{(x+y)}^{3}}-3xy(x+y)}\] \[=\frac{16-2}{64-3\times 4}=\frac{14}{52}=\frac{7}{26}\]You need to login to perform this action.
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