Bài 5:
a: \(A=\sqrt{12-6\sqrt3}+\sqrt{21-12\sqrt3}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt3+3}+\sqrt{12-2\cdot2\sqrt3\cdot3+9}\)
\(=\sqrt{\left(3-\sqrt3\right)^2}+\sqrt{\left(2\sqrt3-3\right)^2}=3-\sqrt3+2\sqrt3-3=\sqrt3\)
b: \(\sqrt{2+\sqrt3}+\sqrt{3-\sqrt5}-\sqrt{\frac52}\)
\(=\frac{1}{\sqrt2}\left(\sqrt{4+2\sqrt3}+\sqrt{6-2\sqrt5}-\sqrt5\right)\)
\(=\frac{1}{\sqrt2}\left(\sqrt3+1+\sqrt5-1-\sqrt5\right)=\frac{\sqrt3}{\sqrt2}\)
\(\sqrt{2-\sqrt3}+\sqrt{3+\sqrt5}-\sqrt{\frac32}\)
\(=\frac{1}{\sqrt2}\left(\sqrt{4-2\sqrt3}+\sqrt{6+2\sqrt5}-\sqrt3\right)\)
\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt3-1\right)^2}+\sqrt{\left(\sqrt5+1\right)^2}-\sqrt3\right)\)
\(=\frac{1}{\sqrt2}\left(\sqrt3-1+\sqrt5+1-\sqrt3\right)=\frac{1}{\sqrt2}\cdot\sqrt5=\sqrt{\frac52}=\frac{\sqrt{10}}{2}\)
Ta có: \(B=5\left(\sqrt{2+\sqrt3}+\sqrt{3-\sqrt5}-\sqrt{\frac52}\right)^2\) +\(\left(\sqrt{2-\sqrt3}+\sqrt{3+\sqrt5}-\sqrt{\frac32}\right)^2\)
\(=5\cdot\frac32+\frac{10}{4}=\frac{15}{2}+\frac52=\frac{20}{2}=10\)
