A=((căn (căn 5 )+2) +(căn (căn 5 )-2) )/(căn( căn 5)+1)

Đặt `A=`$\dfrac{\sqrt{√5+2}+\sqrt{√5-2}}{\sqrt{√5+1}}$ 

`<=> A²=`$\dfrac{\sqrt{5}+2+2\sqrt{5-4}+\sqrt{5}-2}{\sqrt{5}+1}$ 

`=`$\dfrac{2\sqrt{5}+2}{\sqrt{5}+1}$ 

`=`$\dfrac{2(\sqrt{5}+1)}{\sqrt{5}+1}$ 

`=2`

`=> A =√2`

Hay$\dfrac{\sqrt{√5+2}+\sqrt{√5-2}}{\sqrt{√5+1}}$ `=√2`

Đáp án:

`A=\sqrt[2]` 

Giải thích các bước giải:

Đặt `A={\sqrt[\sqrt[5]+2]+\sqrt[\sqrt[5]-2]}/{\sqrt[\sqrt[5]+1]}`

`⇒` `A^2={(\sqrt[\sqrt[5]+2]+\sqrt[\sqrt[5]-2])^2}/{(\sqrt[\sqrt[5]+1])^2}`

`A^2={\sqrt[5]+2+2\sqrt[5-4]+\sqrt[5]-2}/{\sqrt[5]+1}`

`A^2={2\sqrt[5]+2}/{\sqrt[5]+1}`

`A^2={2(\sqrt[5]+1)}/{\sqrt[5]+1}`

`A^2=2`

`⇒` `A=\sqrt[2]`