A) 52 |
B) 76 |
C) 4 |
D) 64 |
A) 52 |
\( \Large x = \frac{\sqrt{3}+1}{\sqrt{3}-1} \)
\( \Large y = \frac{\sqrt{3}-1}{\sqrt{3}+1} \)
\( \Large x + y = \frac{ \left(\sqrt{3}+1\right)^{2} + \left(\sqrt{3}-1\right)^{2} }{ \left(\sqrt{3}-1\right) \left(\sqrt{3}-1\right) } \)
= \( \Large \frac{3+1+2\sqrt{3}+3+1-2\sqrt{3}}{3 - 1} = \frac{8}{2} \)
= x + y = 4
and xy = \( \Large \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}-1}{\sqrt{3}+1} = 1 \)
Therefore, \( \Large \frac{x^{2}}{y}+\frac{y^{2}}{x} = \frac{x^{3} + y^{3}}{xy} \)
= \( \Large \frac{ \left(x+y\right)^{3} - 3xy \left(x+y\right) }{xy} \)
= \( \Large \frac{ \left(4\right)^{3} - 3 \times 1 \times 4 }{1} \)
= \( \Large \frac{64 - 12}{1} = 52 \)