Problem: The fraction 2(2+6)32+3\dfrac{2(\sqrt{2}+\sqrt{6})}{3 \sqrt{2+\sqrt{3}}}32+32(2+6) is equal to
Answer Choices:
A. 223\dfrac{2 \sqrt{2}}{3}322
B. 111
C. 233\dfrac{2 \sqrt{3}}{3}323
D. 43\dfrac{4}{3}34
E. 169\dfrac{16}{9}916 Solution:
Multiplying the numerator and denominator of the given fraction by 2\sqrt{2}2 gives
2(2+6)32+3=4(1+3)34+23=4(1+3)3(1+3)2=43\dfrac{2(\sqrt{2}+\sqrt{6})}{3 \sqrt{2+\sqrt{3}}}=\dfrac{4(1+\sqrt{3})}{3 \sqrt{4+2 \sqrt{3}}}=\dfrac{4(1+\sqrt{3})}{3 \sqrt{(1+\sqrt{3})^{2}}}=\dfrac{4}{3}32+32(2+6)=34+234(1+3)=3(1+3)24(1+3)=34.