\(\frac{4(\sqrt{6} + \sqrt{2})}{\sqrt{6}-\sqrt{2}} - \frac{2+\sqrt{3}}{2-\sqrt{3}} =\frac{4(\sqrt{6} + \sqrt{2})(\sqrt{6}+\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})} - \frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}\)

\(=\frac{4(6 + 2\sqrt{12} + 2)}{6-2} - \frac{4+4\sqrt{3}+3}{4-3}\)

\(=\frac{4(8 + 2\sqrt{12})}{4} - \frac{7+4\sqrt{3}}{1}\)

\(=8 + 2\sqrt{12} - (7+4\sqrt{3})\)

\(=8 + 4\sqrt{3} - 7-4\sqrt{3}\)

\(= 1\)

Answer: A

Cheers,
Brent
_________________

We can rationalize each term as follows:


4(√6 + √2)/(√6 - √2) = 4(√6 + √2)(√6 + √2)/(√6 - √2)(√6 + √2)

= 4[6 + 2 + 2√6√2]/[6 - 2]

= 8 + 4√3


(2 + √3)/(2 - √3) = (2 + √3)(2 + √3)/(2 - √3)(2 + √3)

= [4 + 3 + 2(2)(√3)]/[4 - 3]

= 7 + 4√3

Thus, subtracting, we have: 1

Answer A
_________________

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.