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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 ?

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    $\begingroup$ what is your Problem here`? $\endgroup$ – Dr. Sonnhard Graubner May 24 '17 at 20:18
  • $\begingroup$ @Dr.SonnhardGraubner . how simplify ? $\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} $ $\endgroup$ – Almot1960 May 24 '17 at 20:20
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    $\begingroup$ Hint: subtract denominator from numerator. i.e. what is $x-1$? $\endgroup$ – achille hui May 24 '17 at 20:21
  • $\begingroup$ Can you try long division? $\endgroup$ – swoopin_swallow May 24 '17 at 20:23
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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 $2 = u^2$

$(v + 2u+u^2)+(u+v+2) = u(u+v+2) + (u+v+2) = (u+1)(u+v+2)\\ x = \sqrt 2 +1$

$\frac 1x = \frac 1{1+\sqrt 2} = \sqrt 2-1$

$x + \frac 1x = 2\sqrt 2$

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  • $\begingroup$ but : $x+1/x= 2\sqrt 2$ $\endgroup$ – Almot1960 May 24 '17 at 21:01
  • $\begingroup$ @Almot1960 thanks, I see my mistake and have fixed it. $\endgroup$ – Doug M May 24 '17 at 21:16
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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 $$x+\frac 1x\;=\;2\sqrt 2$$

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

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

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

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

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

$$ 8 - y^2 $$

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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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}$$

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  • $\begingroup$ $\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18}=( \sqrt{2}+1) ( \sqrt{2} + \sqrt{3} + \sqrt{4} )$ but. I can not prove $\endgroup$ – Almot1960 May 24 '17 at 20:26
  • $\begingroup$ you must multiply out These terms $\endgroup$ – Dr. Sonnhard Graubner May 24 '17 at 20:28
  • $\begingroup$ Downvote explained: Insertion of $x$'s value and just printing the final result (known to the OP, cf comment by him) meets IMO the criterion "This answer is not useful." $\endgroup$ – Hanno Jun 1 '17 at 13:24

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