Đáp án:
`1)` `\sqrt{17+12\sqrt{2}}=3+2\sqrt{2}`
`2)` `{\sqrt{2}-\sqrt{11+6\sqrt{2}}}/{\sqrt{6+2\sqrt{5}}-\sqrt{5}}=-3`
`3)` `\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{3}+1`
Giải thích các bước giải:
`1)` `\sqrt{17+12\sqrt{2}}`
`=\sqrt{9+12\sqrt{2}+8}`
`=\sqrt{3^2+2.3.2\sqrt{2}+(2\sqrt{2})^2}`
`=\sqrt{(3+2\sqrt{2})^2}`
`=|3+2\sqrt{2}|=3+2\sqrt{2}`
Vậy: `\sqrt{17+12\sqrt{2}}=3+2\sqrt{2}`
$\\$
`2)` `{\sqrt{2}-\sqrt{11+6\sqrt{2}}}/{\sqrt{6+2\sqrt{5}}-\sqrt{5}}`
`={\sqrt{2}-\sqrt{3^2+2.3\sqrt{2}+2}}/{\sqrt{5+2.\sqrt{5}.1+1^2}-\sqrt{5}}`
`={\sqrt{2}-\sqrt{(3+\sqrt{2})^2}}/{\sqrt{(\sqrt{5}+1)^2}-\sqrt{5}}`
`={\sqrt{2}-|3+\sqrt{2}|}/{|\sqrt{5}+1|-\sqrt{5}}`
`={\sqrt{2}-(3+\sqrt{2})}/{\sqrt{5}+1-\sqrt{5}}`
`={-3}/1=-3`
Vậy: `{\sqrt{2}-\sqrt{11+6\sqrt{2}}}/{\sqrt{6+2\sqrt{5}}-\sqrt{5}}=-3`
$\\$
`3)` `\sqrt{6+2\sqrt{4-2\sqrt{3}}}`
`=\sqrt{6+2\sqrt{3-2.\sqrt{3}.1+1^2}}`
`=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}`
`=\sqrt{6+2|\sqrt{3}-1|}`
`=\sqrt{6+2.(\sqrt{3}-1)}`
`=\sqrt{4+2\sqrt{3}}`
`=\sqrt{3+2.\sqrt{3}.1+1^2}`
`=\sqrt{(\sqrt{3}+1)^2}`
`=|\sqrt{3}+1|=\sqrt{3}+1`
Vậy: `\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{3}+1`
