Question

Tính:

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

Answer 1

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

Answer 2

\(\sqrt{6+2\sqrt{3}\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}=\sqrt{6+2\sqrt{2}\cdot\sqrt{2-\sqrt{3}}}=\sqrt{6+\sqrt{16-8\sqrt{3}}}=\sqrt{6+\sqrt{12-2\cdot2\sqrt{3}\cdot2+4}}=\sqrt{6+\sqrt{\left(2\sqrt{3}-2\right)^2}}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)