So sánh 4 và 2 căn 6 1

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`