<s>[INST] $x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$ 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 ?
 [/INST] Achille hui's comment expanded:  
$$x-1\:=\:\frac{\left(\sqrt 6 +4+3\sqrt 2\right)-\left(\sqrt 2 +2\right)}{\sqrt{3}+\sqrt{2}+2}\:
=\:\frac{\sqrt{2}\sqrt{3}+\sqrt 2\sqrt 2+\sqrt 2\cdot2}{\sqrt{3}+\sqrt{2}+2}\:=\:\sqrt 2 \\[4ex]
\implies\; x=\sqrt 2 +1\quad\implies\;\frac 1x=\sqrt 2-1
$$
which yields
$$x+\frac 1x\;=\;2\sqrt 2$$
</s>