Question

[Vũ Ah minh 32]

Rút gọn biểu thức sau : 

\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)

Answer 1

[Nguyễn Hoàng Bảo Nhi]

Ta có : 

\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)

\(=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-\sqrt{128}}}}\)

Ta có : 

\(18-\sqrt{128}=18-8\sqrt{2}=16-2.4.\sqrt{2}+2=\left(4-\sqrt{2}\right)^2\)

Vậy 

\(\sqrt{18-\sqrt{128}}=4-\sqrt{2}\)

Thay vào ta có

\(\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-\sqrt{128}}}}\)

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

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

Lại có : 

\(4+2\sqrt{3}=3+2.1.\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\)

Do đó : 

\(\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

Vậy : 

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

\(=\sqrt{4-2\sqrt{3}}\)

\(=\sqrt{3-2.1.\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\sqrt{3}-1\)

Vậy : \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}=\sqrt{3}-1\)