The fraction 2(√2+√6)3√2+√3 is equal to

1973 AMC 12 Problem 13

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13.

The fraction

\frac{2(\sqrt2+\sqrt6)}{3\sqrt{2+\sqrt3}}

is equal to

\dfrac{2\sqrt2}{3}

1

\dfrac{2\sqrt3}{3}

\dfrac43

\dfrac{16}{9}

All quantities are positive, so compare the square of the fraction with the squares of the choices

Use

(\sqrt2+\sqrt6)^2=8+4\sqrt3

The fraction is positive, and its square is

\begin{aligned} &\frac{4(\sqrt2+\sqrt6)^2} {9(2+\sqrt3)}\\ &\quad=\frac{4(8+4\sqrt3)} {9(2+\sqrt3)}\\ &\quad=\frac{16(2+\sqrt3)} {9(2+\sqrt3)}\\ &\quad=\frac{16}{9}. \end{aligned}

Therefore the original fraction is

\frac{4}{3}.

Therefore, the correct answer is D.