Answer
$\dfrac{2+3\sqrt 2}{7}$
Work Step by Step
Simplify. $\dfrac{2\sqrt 3+\sqrt 6}{2\sqrt 6+\sqrt 3}$
Now, $\dfrac{2\sqrt 3+\sqrt 6}{2\sqrt 6+\sqrt 3}=\dfrac{(2\sqrt 3+\sqrt 6)(2\sqrt 6-\sqrt 3)}{(2\sqrt 6+\sqrt 3)(2\sqrt 6-\sqrt 3)}$
or, $=\dfrac{4\sqrt{18}-2\sqrt9+2\sqrt{36}-\sqrt{18}}{(2\sqrt 6)^2-(\sqrt 3)^2}$
or, $=\dfrac{6+9\sqrt 2}{24-3}$
or, $=\dfrac{2+3\sqrt 2}{7}$
