a)

\((4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})^2=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})\)

\(=2(4^2-15)=2\)

b)

\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}=\sqrt{(8+2\sqrt{15})+2+2(\sqrt{6}+\sqrt{10})}\)

\(=\sqrt{(\sqrt{5}+\sqrt{3})^2+2\sqrt{2}(\sqrt{3}+\sqrt{5})+2}\)

\(=\sqrt{(\sqrt{5}+\sqrt{3}+\sqrt{2})^2}=\sqrt{5}+\sqrt{3}+\sqrt{2}\)

c)

\((\sqrt{5+2\sqrt{9\sqrt{5}-19}}-\sqrt{7-\sqrt{5}}):(2\sqrt{\sqrt{5}-2})\)

\(=(\sqrt{(5+2\sqrt{9\sqrt{5}-19})(\sqrt{5}+2)}-\sqrt{(7-\sqrt{5})(\sqrt{5}+2)}):(2\sqrt{(\sqrt{5}-2)(\sqrt{5}+2)})\)

\(=[\sqrt{10+5\sqrt{5}+2\sqrt{(9\sqrt{5}-19)(9+4\sqrt{5})}}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{10+5\sqrt{5}+2\sqrt{9+5\sqrt{5}}}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{(9+5\sqrt{5})+2\sqrt{9+5\sqrt{5}}+1}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{(\sqrt{9+5\sqrt{5}}+1)^2}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{9+5\sqrt{5}}+1-\sqrt{9+5\sqrt{5}}]:2=\frac{1}{2}\)

d)

\((\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}})^2=18+2\sqrt{(9+\sqrt{5})(9-\sqrt{5})}=18+4\sqrt{19}\)

\(\Rightarrow \sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}=\sqrt{18+4\sqrt{19}}\)

Do đó:
\(\frac{\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{3-2\sqrt{2}}=\frac{\sqrt{18+4\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{2+1-2\sqrt{2.1}}\)

\(=\frac{\sqrt{2}.\sqrt{9+2\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-(\sqrt{2}-1)=1\)

c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)

b, t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).(\(\sqrt{10}\)-\(\sqrt{2}\))

t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).\(\sqrt{2}\)(\(\sqrt{5}\) -1)

t = (\(\sqrt{5}\) -1).(\(\sqrt{5}\) -1).(3 +\(\sqrt{5}\))

t = (\(\sqrt{5}\) -1)².(3 +\(\sqrt{5}\))

t = (5 - \(2\sqrt{5}\)+1).(3 +\(\sqrt{5}\))

t = 15 + \(5\sqrt{5}\) \(-6\sqrt{5}\)-10+1+\(\sqrt{5}\)

t = 6

a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)

       = \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)

       = \(\frac{-2\sqrt{6}}{2}\)

       = \(-\sqrt{6}\)

a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)

\(=\sqrt{2}-1-3-\sqrt{2}\)

=-4

b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)

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

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

c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)

\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)

\(=3\sqrt{5}-6\)

d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)

\(=\sqrt{7}-2+4-\sqrt{7}+8\)

=10

\(A=\sqrt{9-6\sqrt{7}+7}+\sqrt{3-2\sqrt{21}+7}\)

\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)

\(=3-\sqrt{7}+\sqrt{7}-\sqrt{3}\)

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

\(B=\sqrt{25+2\sqrt{75}+3}+\sqrt{16-2\sqrt{48}+3}\)

\(=\sqrt{\left(5+\sqrt{3}\right)^2}+\sqrt{\left(4-\sqrt{3}\right)^2}\)

\(=5+\sqrt{3}+4-\sqrt{3}\)

\(=9\)

\(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-12\sqrt{5}+9}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}\\ =1\)