`1)sqrt{17-12sqrt2}+sqrt{9+4sqrt2}`
`=sqrt{(3-2sqrt2)^2}+sqrt{(2sqrt2+1)^2}`
`=3-2sqrt2+2sqrt2+1`
`=4`
`2)sqrt{13+30sqrt{2+sqrt{9+4sqrt2}}}`
`=sqrt{13+30sqrt{2+sqrt{(2sqrt2+1)^2}}}`
`=sqrt{13+30sqrt{2+sqrt2+1}}`
`=sqrt{13+30sqrt{3+2sqrt2}}`
`=sqrt{13+30(sqrt2+1)}`
`=sqrt{43+30sqrt2}`
`=sqrt{(5+3sqrt2)^2`
`=5+3sqrt2`
`3)sqrt{sqrt5-sqrt{3-sqrt{29-12sqrt5}}}`
`=sqrt{sqrt5-sqrt{3-sqrt{(3-2sqrt5)^2}}}`
`=sqrt{sqrt5-sqrt{6-2sqrt5}}`
`=sqrt{sqrt5-sqrt{(sqrt5-1)^2}}`
`=sqrt{sqrt5-sqrt5+1}=1`
4) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
Đặt biểu thức là A (A > 0)
Ta có:
A2= \(4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)
= \(8-2\sqrt{16-7}\)
= \(8-6\)
= \(2\)
\(\Leftrightarrow A=\pm\sqrt{2}\) nhưng vì A > 0 nên A = \(\sqrt{2}\)
