\(e,=\left(\sqrt{3}-\sqrt{5}\right)\sqrt{8-2\sqrt{15}}+\sqrt{\left(2+\sqrt{3}\right)^2}\\ =\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)+2+\sqrt{3}\\ =-\left(8-2\sqrt{15}\right)+2+\sqrt{3}=-8+2\sqrt{15}+2+\sqrt{3}=2\sqrt{15}+\sqrt{3}-6\)
\(f,=\left(\sqrt{2}+\sqrt{10}\right)\sqrt{5-\sqrt{5+4\left(\sqrt{5}+1\right)}}\\ =\left(\sqrt{2}+\sqrt{10}\right)\sqrt{5-\sqrt{9+4\sqrt{5}}}\\ =\sqrt{2}\left(1+\sqrt{5}\right)\sqrt{5-\left(2+\sqrt{5}\right)}\\ =\left(1+\sqrt{5}\right)\sqrt{10-4-2\sqrt{5}}=\left(\sqrt{5}+1\right)\sqrt{6-2\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=4\)
e: Ta có: \(\left(\sqrt{6}-\sqrt{10}\right)\cdot\sqrt{4+\sqrt{15}}+\sqrt{7+4\sqrt{3}}\)
\(=\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)+2+\sqrt{3}\)
\(=3-5+2+\sqrt{3}\)
\(=\sqrt{3}\)
f: ta có: \(\left(\sqrt{2}+\sqrt{10}\right)\cdot\sqrt{5-\sqrt{5+4\sqrt{6+2\sqrt{5}}}}\)
\(=\left(\sqrt{10}+\sqrt{2}\right)\cdot\sqrt{5-\sqrt{9+4\sqrt{5}}}\)
\(=\left(\sqrt{5}+1\right)\cdot\sqrt{2}\cdot\sqrt{3-\sqrt{5}}\)
=4
