$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$ then $x+\dfrac{1}{x}=?$ 
let :
  $$x = \frac{{\sqrt 3  + \sqrt 6  + \sqrt {16}  + \sqrt {18} }}{{\sqrt 2  + \sqrt 3  + \sqrt 4 }}$$
then :
$$x+\dfrac{1}{x}=?$$

My try :
$$\sqrt 3  + \sqrt 6  + \sqrt {16}  + \sqrt {18}=\sqrt3+\sqrt 3\times\sqrt2+4+3\sqrt2 \\ \sqrt3(1+\sqrt2+3)+4$$
Now what ?

simplify the term $$\frac{2+\sqrt{2}+\sqrt{3}}{4+3 \sqrt{2}+\sqrt{3}+\sqrt{6}}+\frac{4+3
   \sqrt{2}+\sqrt{3}+\sqrt{6}}{2+\sqrt{2}+\sqrt{3}}$$
$$\sqrt{16}=4,\sqrt{18}=3\sqrt{2}$$
the final result is $2\sqrt{2}$$
