giải giùm em bài ni nhanh lên ạ: √2-√3 - √2+ √3

`A=` $\sqrt{2-\sqrt{3}}$ `-`$\sqrt{2+\sqrt{3}}$

`=>A^2=` ${2-\sqrt{3}}$ `-2` $\sqrt{({2-\sqrt{3}})({2+\sqrt{3})}}$ `+`${2+\sqrt{3}}$

`=>A^2=4-2`

`=>A^2=2`

`=>A=` $±\sqrt{2}$ 

Đáp án+Giải thích các bước giải:

`\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}`

`=\sqrt{\frac{4-2\sqrt{3}}{2}}-\sqrt{\frac{4+2\sqrt{3}}{2}}`

`=\sqrt{\frac{3+1-2\sqrt{3}}{2}}-\sqrt{\frac{3+1+2\sqrt{3}}{2}}`

`=\sqrt{\frac{(\sqrt{3})^2-2.\sqrt{3}.1+1^2}{2}}-\sqrt{\frac{(\sqrt{3})^2+2\sqrt{3}.1+1^2}{2}}`

`=\sqrt{\frac{(\sqrt{3}-1)^2}{2}}-\sqrt{\frac{(\sqrt{3}+1)^2}{2}}`

`=\frac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{2}}-\frac{\sqrt{(\sqrt{3}+1)^2}}{\sqrt{2}}`

`=\frac{|\sqrt{3}-1|}{\sqrt{2}}-\frac{|\sqrt{3}+1|}{\sqrt{2}}`

`=\frac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}`

`=\frac{-2}{\sqrt{2}}`

`=-\sqrt{2}`