\(a=\sqrt{3+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}+\sqrt{3-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}\)

\(a=\sqrt{3+\sqrt{3}+\sqrt{2}}+\sqrt{3-\sqrt{3}-\sqrt{2}}\)

\(\Rightarrow a^2=3+\sqrt{3}+\sqrt{2}+3-\sqrt{3}-\sqrt{2}+2\sqrt{\left(3+\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{3}-\sqrt{2}\right)}\)\(\Rightarrow VT=3+\sqrt{3}+\sqrt{2}+3-\sqrt{3}-\sqrt{2}+2\sqrt{\left(3+\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{3}-\sqrt{2}\right)}-2\sqrt{\left(3+\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{3}-\sqrt{2}\right)}-2\)

\(=6-2=4\) ??? đề bài có sai ko bn?

\(a^2=6+2\sqrt{9-\left(5+2\sqrt{3}\right)}=6+2\sqrt{4-2\sqrt{3}}=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)

\(\Rightarrow a=\sqrt{3}+1\)

\(\Rightarrow a^2-2a-2=\left(a-1\right)^2-3=\left(\sqrt{3}+1-1\right)^2-3=3-3=0\)

Lời giải:

Ta có:

$a^2=3+\sqrt{5+2\sqrt{3}}+3-\sqrt{5+2\sqrt{3}}+2\sqrt{(3+\sqrt{5+2\sqrt{3}})(3-\sqrt{5+2\sqrt{3}})}$

$=6+2\sqrt{3^2-(5+2\sqrt{3})}=6+2\sqrt{4-2\sqrt{3}}=6+2\sqrt{3+1-2\sqrt{3}}$

$=6+2\sqrt{(\sqrt{3}-1)^2}=6+2(\sqrt{3}-1)=4+2\sqrt{3}=(\sqrt{3}+1)^2$

$\Rightarrow a=\sqrt{3}+1$ (do $a\geq 0$)

Do đó:

$a^2-2a-2=4+2\sqrt{3}-2(\sqrt{3}+1)-2=0$ (đpcm)

ta có : \(a=\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}}\)

\(\Rightarrow a^2=6+2\sqrt{4-2\sqrt{3}}=6+2\sqrt{\left(\sqrt{3}-1\right)^2}=4+2\sqrt{3}\)

\(\Rightarrow a=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\) (do \(a>0\) )

\(\Rightarrow a^2-2a-2=4+2\sqrt{3}-2\left(\sqrt{3}+1\right)-2=0\)

Mysterious Person giúp mk nha

Ta có a² = 6 + 2\(\sqrt{4-2\sqrt{3}}\)= 6 + \(2\sqrt{3}\)- 2 = 4 + 2\(\sqrt{3}\)= (\(\sqrt{3}\)+ 1)²

=> a = \(1+\sqrt{3}\)

Từ đó => a2​- 2a - 2 = 0

Cái đề bạn bị sai rồi nhé