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I have tried breaking up the fraction into two parts so the first part would cancel to 1, but to no avail. Rewriting the expression also doesn't seem to work, so how can I start?

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  • $\begingroup$ One observation: The denominator is of the form $a\sqrt2+b\sqrt3$ for some $a, b.$ $\endgroup$
    – Bumblebee
    Nov 4, 2017 at 23:56

2 Answers 2

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It is clear that you cannot manipulate numerator, but for denominator:

$\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4 =\\ \sqrt{2} + \sqrt{3}+\sqrt{2}\sqrt{3}+\sqrt{2}\sqrt{4}+2\sqrt{4}=\\ \sqrt{2}+\sqrt{3}+\sqrt{4} + \sqrt{2}(\sqrt{3} + \sqrt{4} + \sqrt{2})$

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Let $\frac {\sqrt2+\sqrt3+\sqrt4}{\sqrt2+\sqrt3+\sqrt6+\sqrt8+4} = \frac{x}{y}$. By componendo and dividendo (Brilliant):

$$\frac {\sqrt2+\sqrt3+\sqrt6+\sqrt8+4}{\sqrt2+\sqrt3+\sqrt4} = \frac{y}{x}$$ $$\Rightarrow \frac {\sqrt2+\sqrt3+\sqrt6+\sqrt8+4 -(\sqrt{2} + \sqrt{3} + \sqrt{4})}{\sqrt2+\sqrt3+\sqrt4} = \frac{y-x}{x}$$ $$\Rightarrow \frac {\sqrt4 + \sqrt6+\sqrt8}{\sqrt2+\sqrt3+\sqrt4} = \frac{y-x}{x}$$ $$\Rightarrow \frac {\sqrt{2} (\sqrt 2 + \sqrt3+\sqrt4)}{\sqrt2+\sqrt3+\sqrt4} = \frac{y}{x} - 1$$ $$\Rightarrow 1 + \sqrt{2} = \frac{y}{x} \Rightarrow \frac{x}{y} = \frac{1}{1 + \sqrt2} = \sqrt{2} - 1$$

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    $\begingroup$ So ${x\over y}=\sqrt{2}-1$, brilliant! $\endgroup$
    – nonuser
    Aug 24, 2020 at 6:37

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