Đáp án:
`a)` `Q=-\sqrt{2}`
`b)` `-\sqrt{6}`
`c)` `2\sqrt{3}`
Giải thích các bước giải:
`q)` `Q=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}`
`=>\sqrt{2} Q=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}`
`=\sqrt{3-2.\sqrt{3}.1+1}-\sqrt{3+2.\sqrt{3}.1+1}`
`=\sqrt{(\sqrt{3}-1)^2}-\sqrt{(\sqrt{3}+1)^2}`
`=|\sqrt{3}-1|-|\sqrt{3}+1|`
`=\sqrt{3}-1-(\sqrt{3}+1)`
`=-2`
`=>\sqrt{2}Q=-2`
`=>Q={-2}/{\sqrt{2}=-\sqrt{2}`
Vậy `Q=-\sqrt{2}`
$\\$
`r)` `R=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}`
`=>R^2=(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}})^2`
`=>R^2=12-3\sqrt{7}-2\sqrt{(12-3\sqrt{7}).(12+\sqrt{7})}+12+3\sqrt{7}`
`=>R^2=24-2\sqrt{12^2-(3\sqrt{7})^2}`
`=>R^2=24-2.\sqrt{144-63}=24-2.9=6` $(1)$
Vì `R=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}<0`
`=>R<0`
Từ `(1)=>R=-\sqrt{6}`
Vậy `\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}=-\sqrt{6}`
$\\$
`s)` `\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}`
`\sqrt{5-\sqrt{(2\sqrt{3})^2+2.2\sqrt{3}.1+1}}+\sqrt{3+\sqrt{(2\sqrt{3})^2+2.2\sqrt{3}.1+1}}`
`=\sqrt{5-\sqrt{(2\sqrt{3}+1)^2}}+\sqrt{3+\sqrt{(2\sqrt{3}+1)^2}}`
`=\sqrt{5-|2\sqrt{3}+1|}+\sqrt{3+|2\sqrt{3}+1|}`
`=\sqrt{5-(2\sqrt{3}+1)}+\sqrt{3+2\sqrt{3}+1}`
`=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}`
`=\sqrt{3-2.\sqrt{3}.1+1}+\sqrt{3+2.\sqrt{3}.1+1}`
`=\sqrt{(\sqrt{3}-1)^2}+\sqrt{(\sqrt{3}+1)^2}`
`=|\sqrt{3}-1|+|\sqrt{3}+1|`
`=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}`
Vậy: `\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}=2\sqrt{3}`
