<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] (Too long for a comment.)  For a more systematic, albeit grossly overkill in this case, approach:


*

*with $\,u=\sqrt{2}\,$ and $\,v=\sqrt{3}\,$:


$$
x = \frac{{\sqrt 3  + \sqrt 6  + 4 + 3 \sqrt {2} }}{{\sqrt 2  + \sqrt 3  + 2 }} = \frac{4+ 3u + v + uv}{2+u+v}
$$


*

*eliminate $u$ by Resultant[x (2 + u + v) - (4 + 3 u + v + u v), u^2 - 2, u]:


$$
v^2 x^2 - 2 v^2 x - v^2 + 4 v x^2 - 8 v x - 4 v + 2 x^2 - 4 x - 2
$$


*

*then $v$ by Resultant[v^2 x^2 - 2 v^2 x - v^2 + 4 v x^2 - 8 v x - 4 v + 2 x^2 - 4 x - 2, v^2 - 3, v]:


$$
-23 x^4 + 92 x^3 - 46 x^2 - 92 x - 23 = -23 (x^2 - 2 x - 1)^2
$$


*

*then let $y = x + 1/x \iff x^2-xy+1=0\,$, and Resultant[x^2 - 2 x - 1, x^2 - x y + 1, x]:


$$
8 - y^2
$$
Therefore $x+1/x=y=\pm \sqrt{8}\,$ and, since $\,x\,$ is known to be positive, $\,x+1/x=+\sqrt{8}=2\sqrt{2}\,$.
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