Question

a,\(\sqrt{1+2\sqrt{2}+\sqrt{11+6\sqrt{2}}}\)

b,\(\sqrt{10-2\sqrt{21}}+\sqrt{4+2\sqrt{3}}\)

c,\(\sqrt{1+\dfrac{\sqrt{3}}{2}}+\sqrt{1-\dfrac{\sqrt{3}}{2}}\)

d,\(\sqrt{15+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\)

Answer 1

\(a.\sqrt{1+2\sqrt{2}+\sqrt{11+6\sqrt{2}}}=\sqrt{1+2\sqrt{2}+\sqrt{9+2.3\sqrt{2}+2}}=\sqrt{1+2\sqrt{2}+3+\sqrt{2}}=\sqrt{4+3\sqrt{2}}\)

\(b.\sqrt{10-2\sqrt{21}}+\sqrt{4+2\sqrt{3}}=\sqrt{7-2\sqrt{7}.\sqrt{3}+3}+\sqrt{3+2\sqrt{3}+1}=\sqrt{7}-\sqrt{3}+\sqrt{3}+1=\sqrt{7}+1\)

\(c.\sqrt{1+\dfrac{\sqrt{3}}{2}}+\sqrt{1-\dfrac{\sqrt{3}}{2}}=\sqrt{\dfrac{3}{4}+2.\dfrac{\sqrt{3}}{2}.\dfrac{1}{2}+\dfrac{1}{4}}+\sqrt{\dfrac{3}{4}-2.\dfrac{\sqrt{3}}{2}.\dfrac{1}{2}+\dfrac{1}{4}}=\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}=\sqrt{3}\)

\(d.\sqrt{15+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}=\sqrt{9+2.3\sqrt{6}+6}-\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3+\sqrt{6}-3\sqrt{2}+\sqrt{3}=\sqrt{3}\left(\sqrt{3}+\sqrt{2}-\sqrt{6}+1\right)\)