0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ One observation: The denominator is of the form $a\sqrt2+b\sqrt3$ for some $a, b.$ $\endgroup$
    – Bumblebee
    Nov 4 '17 at 23:56
5
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
1
  • 1
    $\begingroup$ So ${x\over y}=\sqrt{2}-1$, brilliant! $\endgroup$
    – Aqua
    Aug 24 '20 at 6:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.