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: `11-\sqrt{3}\sqrt{2}.[1-\sqrt{3}]`

`\qquad ` [do `1-\sqrt{3} 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`