$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$ 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 ?
 A: Suppose we say $u = \sqrt 2, v = \sqrt 3$
then 
$\sqrt 2 + \sqrt 3 + \sqrt 4 = u+v + 2$
and $(\sqrt 3 +\sqrt 6 + \sqrt 16 + \sqrt 18) = (v+ uv +  3u + 4) = (uv + 2u+2)+(u+v+2)$
in the first term on the right hand side say $2 = u^2$
$(v + 2u+u^2)+(u+v+2) = u(u+v+2) + (u+v+2) = (u+1)(u+v+2)\\
x = \sqrt 2 +1$
$\frac 1x = \frac 1{1+\sqrt 2} = \sqrt 2-1$
$x + \frac 1x = 2\sqrt 2$
A: 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
$$x+\frac 1x\;=\;2\sqrt 2$$
A: Note that
$$\begin{align}
(1+\sqrt2)(2+\sqrt2+\sqrt3)
&=(2+\sqrt2+\sqrt3)+(2\sqrt2+2+\sqrt6)\\\\
&=4+3\sqrt2+\sqrt3+\sqrt6\\\\
&=\sqrt{16}+\sqrt{18}+\sqrt3+\sqrt6
\end{align}$$
Thus
$$x={\sqrt3+\sqrt6+\sqrt{16}+\sqrt{18}\over\sqrt2+\sqrt3+\sqrt4}={(1+\sqrt2)(2+\sqrt2+\sqrt3)\over\sqrt2+\sqrt3+2}=1+\sqrt2$$
so that
$${1\over x}={1\over1+\sqrt2}={1-\sqrt2\over1-2}=\sqrt2-1$$
and thus
$$x+{1\over x}=2\sqrt2$$
Remark:  The first part of the derivation, starting with the factorization, is, needless to say, presented without all the finagling that went into finding it.
A: (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, $\,x+1/x=+\sqrt{8}=2\sqrt{2}\,$.
A: 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}$$
