a) Ta có: \(\left(4+\sqrt{5}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{5}\right)\cdot\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\left(4+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{5}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{5}-2\sqrt{75}\)
b) Ta có: \(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\left(\sqrt{5}-1\right)\cdot\left(3+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-1\right)^2\cdot\left(\sqrt{5}+3\right)\)
\(=\left(6-2\sqrt{5}\right)\cdot\left(\sqrt{5}+3\right)\)
\(=2\cdot\left(3-\sqrt{5}\right)\cdot\left(3+\sqrt{5}\right)\)
\(=2\cdot\left(9-5\right)=2\cdot4=8\)
