Đáp án:
Đặt `A=\frac{\sqrt{2+\sqrt3}}{\sqrt{2-\sqrt3}}`
`=>\sqrt2A=\frac{\sqrt{4+2\sqrt3}}{\sqrt{2-\sqrt3}}`
`=>\sqrt2A=\frac{\sqrt{3+2\sqrt3+1}}{\sqrt{2-\sqrt3}}`
`=>\sqrt2A=\frac{\sqrt{(\sqrt3+1^2)}}{\sqrt{2-\sqrt3}}`
`=>\sqrt2A=\frac{\sqrt{3}+1}{\sqrt{2-\sqrt3}}`
`=>\sqrt2A=\frac{(\sqrt{3}+1)(\sqrt2+\sqrt3)}{2-3}`
`=>\sqrt2A=-(\sqrt{3}+1)(\sqrt2+\sqrt3)`
`=>\sqrt2A=-(3+\sqrt6+\sqrt3+\sqrt2)`
`=>A=-(3sqrt2+2\sqrt3+\sqrt6+2)/2`
