Question

tính giá trị biểu thức

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

b, \(\sqrt{12-6\sqrt{3}}\)

c, \(\sqrt{6-4\sqrt{2}+\sqrt{22-12\sqrt{2}}}\)

Answer 1

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

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

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

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

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

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

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

b) \(\sqrt{12-6\sqrt{3}}\)

\(=\sqrt{9-2\cdot3\cdot\sqrt{3}+3}\)

\(=\sqrt{\left(\sqrt{9}-\sqrt{3}\right)^2}\)

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

c) \(\sqrt{6-4\sqrt{2}+\sqrt{22-12\sqrt{2}}}\)

\(=\sqrt{6-4\sqrt{2}+\sqrt{\left(\sqrt{18}-2\right)^2}}\)

\(=\sqrt{6-4\sqrt{2}+\sqrt{18}-2}\)

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

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