Câu trả lời:

\(\sqrt{6+2\sqrt{4-\sqrt{12}}=}\sqrt{6+2\sqrt{\sqrt{3^2}-2\sqrt{3}\cdot1}+1}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

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