1.85 The expression $\frac{2(\sqrt{2}+\sqrt{6})}{3 \sqrt{2+\sqrt{3}}}$ equals
(A) $\frac{2 \sqrt{2}}{3}$.
(B) 1 .
(C) $\frac{2 \sqrt{3}}{3}$.
(D) $\frac{4}{3}$.
(E) $\frac{16}{9}$.
(24th American High School Mathematics Examination, 1973)
[Sol] Multiplying the numerator and denominator of the fraction by $\sqrt{2}$, we get
$$
\frac{2(\sqrt{2}+\sqrt{6})}{3 \sqrt{2+\sqrt{3}}}=\frac{4(1+\sqrt{3})}{3 \sqrt{4+2 \sqrt{3}}}=\frac{4(1+\sqrt{3})}{3 \sqrt{(1+\sqrt{3})^{2}}}=\frac{4}{3} \text {. }
$$

Therefore, the correct choice is $(D)$.