Đáp án:
`dd)` `\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=-\sqrt{2}`
`ff)` `\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\sqrt{2}`
Giải thích các bước giải:
`dd)` ` \sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}`
`=\sqrt{{8-2\sqrt{7}}/2}-\sqrt{{8+2\sqrt{7}}/2}`
`=\sqrt{{7-2.\sqrt{7}.1+1^2}/2}-\sqrt{{7+2.\sqrt{7}.1+1^2}/2}`
`=\sqrt{{(\sqrt{7}-1)^2}/2}-\sqrt{{(\sqrt{7}+1)^2}/2}`
`=|{\sqrt{7}-1}/\sqrt{2}|-|{\sqrt{7}+1}/\sqrt{2}|`
`={\sqrt{7}-1}/\sqrt{2}-{\sqrt{7}+1}/\sqrt{2}`
`={\sqrt{7}-1-\sqrt{7}-1}/\sqrt{2}`
`=-2/\sqrt{2}=-\sqrt{2}`
Vậy: `\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=-\sqrt{2}`
$\\$
`ff)` `\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}`
`=\sqrt{{4+2\sqrt{3}}/2}-\sqrt{{4-2\sqrt{3}}/2}`
`=\sqrt{{3+2.\sqrt{3}.1+1^2}/2}-\sqrt{{3-2.\sqrt{3}.1+1^2}/2}`
`=\sqrt{{(\sqrt{3}+1)^2}/2}-\sqrt{{(\sqrt{3}-1)^2}/2}`
`=|{\sqrt{3}+1}/\sqrt{2}|-|{\sqrt{3}-1}/\sqrt{2}|`
`={\sqrt{3}+1}/\sqrt{2}-{\sqrt{3}-1}/\sqrt{2}`
`={\sqrt{3}+1-\sqrt{3}+1}/\sqrt{2}`
`=2/\sqrt{2}=\sqrt{2}`
Vậy: `\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\sqrt{2}`
