Lời giải:
a)
\(=\sqrt{\frac{4-2\sqrt{3}}{4}}+\frac{1-\sqrt{3}}{2}=\sqrt{\frac{(\sqrt{3}-1)^2}{2^2}}+\frac{1-\sqrt{3}}{2}\)
\(=\frac{\sqrt{3}-1}{2}+\frac{1-\sqrt{3}}{2}=0\)
b)
\(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}=\sqrt{2+1+2\sqrt{2}}+\sqrt{2^2+2-2.2\sqrt{2}}\)
\(=\sqrt{(\sqrt{2}+1)^2}+\sqrt{(2-\sqrt{2})^2}=\sqrt{2}+1+2-\sqrt{2}=3\)
c)
\(\sqrt{2+\sqrt{3}}=\sqrt{\frac{4+2\sqrt{3}}{2}}=\sqrt{\frac{(\sqrt{3}+1)^2}{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}\)
Do đó, biểu thức đã cho bằng:
\(\frac{\sqrt{3}+1}{2\sqrt{2}}:\left(\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{3}+1}{2\sqrt{6}}\right)\)
\(=\frac{\sqrt{3}+1}{2\sqrt{2}}:\left(\frac{3+\sqrt{3}}{2\sqrt{6}}-\frac{4}{2\sqrt{6}}+\frac{\sqrt{3}+1}{2\sqrt{6}}\right)=\frac{\sqrt{3}+1}{2\sqrt{2}}:\frac{2\sqrt{3}}{2\sqrt{6}}=\frac{\sqrt{3}+1}{2\sqrt{2}}.\sqrt{2}=\frac{\sqrt{3}+1}{2}\)
