<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] 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 $2 = u^2$
$(v + 2u+u^2)+(u+v+2) = u(u+v+2) + (u+v+2) = (u+1)(u+v+2)\\
x = \sqrt 2 +1$
$\frac 1x = \frac 1{1+\sqrt 2} = \sqrt 2-1$
$x + \frac 1x = 2\sqrt 2$
</s>