<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] 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.
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