Đặt `A=\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}`
Vì `\sqrt{\sqrt{2}+1}>\sqrt{\sqrt{2}-1}=>A>0`
Ta có:
`A^2=(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1})^2`
`=\sqrt{2}+1-2\sqrt{(\sqrt{2}+1)(\sqrt{2}-1)}+\sqrt{2}-1`
`=2\sqrt{2}-2\sqrt{2-1^2}=2\sqrt{2}-2`
`=2(\sqrt{2}-1)`
`=>A=\sqrt{2(\sqrt{2}-1)}` (vì `A>0)`
Vậy: `\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2(\sqrt{2}-1)}`(đpcm)
