Biến đổi mẫu số:
\(4+\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}\)
= \(2+2+\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}\)
= \(\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}\)
= \(\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)\)
= \(\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{2.2}+\sqrt{2.3}+\sqrt{2.4}\right)\)
= \(\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\)
= \(\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)\)
Do đó ta có:
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{4+\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}}\)
= \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}\)
= \(\dfrac{1}{1+\sqrt{2}}\)
