Đáp án:
` {2+\sqrt{3}}/{\sqrt{2}+\sqrt{2+\sqrt{3}}}+{2-\sqrt{3}}/{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}`
Giải thích các bước giải:
Ta có:
`\qquad \sqrt{2+\sqrt{3}}=\sqrt{{4+2\sqrt{3}}/2}`
`=\sqrt{{3+2\sqrt{3}.1+1^2}/2}`
`=\sqrt{{(\sqrt{3}+1)^2}/2}=|{\sqrt{3}+1}/\sqrt{2}|`
`={\sqrt{3}+1}/\sqrt{2}`
$\\$
`\qquad \sqrt{2-\sqrt{3}}=\sqrt{{4-2\sqrt{3}}/2}`
`=\sqrt{{3-2\sqrt{3}.1+1^2}/2}`
`=\sqrt{{(\sqrt{3}-1)^2}/2}=|{\sqrt{3}-1}/\sqrt{2}|`
`={\sqrt{3}-1}/\sqrt{2}`
$\\$
`\qquad {2+\sqrt{3}}/{\sqrt{2}+\sqrt{2+\sqrt{3}}}+{2-\sqrt{3}}/{\sqrt{2}-\sqrt{2-\sqrt{3}}}`
`={(2+\sqrt{3})(\sqrt{2}-\sqrt{2+\sqrt{3}})}/{(\sqrt{2}+\sqrt{2+\sqrt{3}}).(\sqrt{2}-\sqrt{2+\sqrt{3}})}+{(2-\sqrt{3}).(\sqrt{2}+\sqrt{2-\sqrt{3}})}/{(\sqrt{2}-\sqrt{2-\sqrt{3}}).(\sqrt{2}+\sqrt{2-\sqrt{3}})}`
`={2\sqrt{2}-2\sqrt{2+\sqrt{3}}+\sqrt{6}-\sqrt{3}.\sqrt{2+\sqrt{3}}}/{2-(2+\sqrt{3})}+{2\sqrt{2}+2\sqrt{2-\sqrt{3}}-\sqrt{6}-\sqrt{3}.\sqrt{2-\sqrt{3}}}/{2-(2-\sqrt{3})}`
`={-2\sqrt{2}+2\sqrt{2+\sqrt{3}}-\sqrt{6}+\sqrt{3}.\sqrt{2+\sqrt{3}}+2\sqrt{2}+2\sqrt{2-\sqrt{3}}-\sqrt{6}-\sqrt{3}.\sqrt{2-\sqrt{3}}}/{\sqrt{3}}`
`={-2\sqrt{6}+2.(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})+\sqrt{3}.(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}})}/{\sqrt{3}}`
`={-2\sqrt{6}}/\sqrt{3}+ 1/\sqrt{3} . [2.({\sqrt{3}+1}/\sqrt{2}+{\sqrt{3}-1}/\sqrt{2})+\sqrt{3}. ({\sqrt{3}+1}/\sqrt{2}-{\sqrt{3}-1}/\sqrt{2})]`
`=-2\sqrt{2}+1/\sqrt{3}. (2. {2\sqrt{3}}/\sqrt{2}+\sqrt{3}. 2/\sqrt{2})`
`=-2\sqrt{2}+1/\sqrt{3}.(2\sqrt{6}+\sqrt{6})`
`=-2\sqrt{2}+3\sqrt{2}=\sqrt{2}`
Vậy: `{2+\sqrt{3}}/{\sqrt{2}+\sqrt{2+\sqrt{3}}}+{2-\sqrt{3}}/{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}`
