Đáp án:
`1)` `1-\sqrt{3}> \sqrt{2}-\sqrt{6}`
`2)` `\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}=0`
Giải thích các bước giải:
`1)` Ta có: `1<\sqrt{3}=>1-\sqrt{3}<0`
Vì `1<\sqrt{2}`
`=>1.(1-\sqrt{3})>\sqrt{2}.(1-\sqrt{3})`
`\qquad ` (do `1-\sqrt{3}<0)`
`=> 1-\sqrt{3}> \sqrt{2}-\sqrt{6}`
$\\$
`2)` `\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}`
`=\sqrt{{8+2\sqrt{7}}/2}-\sqrt{{8-2\sqrt{7}}/2}-\sqrt{2}`
`=\sqrt{{7+2\sqrt{7}.1+1^2}/2}-\sqrt{{7-2\sqrt{7}.1+1^2}/2}-\sqrt{2}`
`=\sqrt{{(\sqrt{7}+1)^2}/2}-\sqrt{{(\sqrt{7}-1)^2}/2}-\sqrt{2}`
`=|{\sqrt{7}+1}/\sqrt{2}|-|{\sqrt{7}-1}/\sqrt{2}|-\sqrt{2}`
`={\sqrt{7}+1-(\sqrt{7}-1)}/\sqrt{2}-\sqrt{2}`
`=2/\sqrt{2}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0`
Vậy: `\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}=0`
