$\sqrt{1+\sqrt{3}+1}$`+`$\sqrt{1-\sqrt{3}+1}$

`#tnvt`

`\sqrt{1+\sqrt{3}+1]+\sqrt{1-\sqrt{3}+1}`

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

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

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

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

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

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

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

`=\sqrt{6}`

Đáp án+Giải thích các bước giải:

 Đặt `A=sqrt{1+sqrt3+1}+sqrt{1-sqrt3+1}`

          `=sqrt{2+sqrt3}+sqrt{2-sqrt3}`

  `=>Asqrt2=sqrt2.(sqrt{2+sqrt3}+sqrt{2-sqrt3})`

       `Asqrt2=sqrt{4+2sqrt3}+sqrt{4-2sqrt3}`

       `Asqrt2=sqrt{3+2sqrt3+1}+sqrt{3-2sqrt3+1}`

       `Asqrt2=sqrt{(sqrt3+1)^2}+sqrt{(sqrt3-1)^2}`

       `Asqrt2=|sqrt3+1|+|sqrt3-1|`

       `Asqrt2=sqrt3+1+sqrt3-1`

       `Asqrt2=2sqrt3`

  `=>A=(2sqrt3)/sqrt2`

         `=sqrt6`

 Vậy `sqrt{1+sqrt3+1}+sqrt{1-sqrt3+1}=sqrt6`