Đáp án + Giải thích các bước giải:
Sửa đề: `\sqrt{2+\sqrt3}+\sqrt{2-\sqrt3}`
Ta có: `(\sqrt{2+\sqrt3}+\sqrt{2-\sqrt3})^2`
`=(\sqrt{(4+2\sqrt3)/2}+\sqrt{(4-2\sqrt3)/2})^2`
`=(\sqrt{(3+2\sqrt3+1)/2}+\sqrt{(3-2\sqrt3+1)/2})^2`
`=(\sqrt{(\sqrt3+1)^2/2}+\sqrt{(\sqrt3-1)^2/2})^2`
`=((|\sqrt3+1|)/\sqrt2+(|\sqrt3-1|)/\sqrt2)^2`
`=((\sqrt3+1+\sqrt3-1)/(\sqrt2))^2`
`=((2\sqrt3)/(\sqrt2))^2`
`=(\sqrt6)^2`
`=6`
`=3+3`
`=3+\sqrt9`
Lại có: `(\sqrt2+1)^2`
`=2+2\sqrt2+1`
`=3+2\sqrt2`
`=3+\sqrt8`
Vì `\sqrt8<\sqrt9`
`=>3+\sqrt8<3+\sqrt9`
`=>(\sqrt2+1)^2<(\sqrt{2+\sqrt3}+\sqrt{2-\sqrt3})^2`
`=>\sqrt2+1<\sqrt{2+\sqrt3}+\sqrt{2-\sqrt3}`
