Đáp án:
`1+\sqrt{2}`
Giải thích các bước giải:
Ta có:
`\qquad {\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}/{\sqrt{2}+\sqrt{3}+\sqrt{4}}`
`= {\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}/{\sqrt{2}+\sqrt{3}+\sqrt{4}}`
`={(\sqrt{2}+\sqrt{3}+2)+(2+\sqrt{6}+\sqrt{8})}/{\sqrt{2}+\sqrt{3}+\sqrt{4}}`
`={(\sqrt{2}+\sqrt{3}+\sqrt{4})+(\sqrt{4}+\sqrt{6}+\sqrt{8})}/{\sqrt{2}+\sqrt{3}+\sqrt{4}}`
`={(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}/{\sqrt{2}+\sqrt{3}+\sqrt{4}}`
`={(\sqrt{2}+\sqrt{3}+\sqrt{4}).(1+\sqrt{2})}/{\sqrt{2}+\sqrt{3}+\sqrt{4}}`
`=1+\sqrt{2}`
Vậy:
`{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}/{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}`
