Ta có: \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
= \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2\cdot\sqrt{2}\cdot\sqrt{16}+2}}}\)
=\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(\sqrt{16}-\sqrt{2}\right)^2}}}\)
=\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16}-\sqrt{2}}}\)
=\(\sqrt{6-2\sqrt{4+\sqrt{12}}}\)
=\(\sqrt{6-2\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
=\(\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{6-2\sqrt{3}-2}\)
=\(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}\cdot1+1}=\sqrt{\left(\sqrt{3}-1\right)^2}\)=\(\sqrt{3}-1\)
\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(16-\sqrt{2}\right)^2}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2+\sqrt{12}+16-\sqrt{2}}}}\)
\(=\sqrt{6-2\sqrt{16+\sqrt{12}}}\) \(=\sqrt{6-2\sqrt{16+2\sqrt{3}}}\)
Thật xin lỗi! Phân tích đến đây là mk tịt r! Bn đok qa có khi lại nghĩ ra đấy!
