Question

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Answer 1

If you want to rationalize the denominator, multiply the fraction by \(\Large \frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}-\sqrt{6}}\) to get...


\[\Large \frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}+\sqrt{6}}\]

\[\Large \frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}+\sqrt{6}}\times\frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}-\sqrt{6}}\]

\[\Large \frac{(\sqrt{2}-\sqrt{6})(\sqrt{2}-\sqrt{6})}{(\sqrt{2}+\sqrt{6})(\sqrt{2}-\sqrt{6})}\]

\[\Large \frac{(\sqrt{2}-\sqrt{6})(\sqrt{2}-\sqrt{6})}{(\sqrt{2})^2-(\sqrt{6})^2}\]

\[\Large \frac{(\sqrt{2}-\sqrt{6})(\sqrt{2}-\sqrt{6})}{2-6}\]

\[\Large \frac{(\sqrt{2}-\sqrt{6})(\sqrt{2}-\sqrt{6})}{-4}\]

I'll let you finish.

Answer 2

jim_thompson5910 san is right!

When I continue calculating the problem
Answer is
\[\sqrt{3} - 2\]