a: \(\left(\sqrt3+2\sqrt5\right)^2-\sqrt{240}\)
\(=3+20+2\cdot\sqrt3\cdot2\sqrt5-4\sqrt{15}\)
\(=23+4\sqrt{15}-4\sqrt{15}=23\)
b: \(\sqrt{3-\sqrt5}\cdot\sqrt{3+\sqrt5}\)
\(=\sqrt{\left(3-\sqrt5\right)\left(3+\sqrt5\right)}\)
\(=\sqrt{9-5}=\sqrt4=2\)
c: \(\frac{4}{\sqrt3+1}+\frac{1}{\sqrt3-2}+\frac{6}{\sqrt3-3}\)
\(=\frac{4\left(\sqrt3-1\right)}{\left(\sqrt3+1\right)\left(\sqrt3-1\right)}-\frac{2+\sqrt3}{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}-\frac{6\left(3+\sqrt3\right)}{\left(3-\sqrt3\right)\left(3+\sqrt3\right)}\)
\(=2\left(\sqrt3-1\right)-2-\sqrt3-\left(3+\sqrt3\right)\)
\(=2\sqrt3-2-2-\sqrt3-3-\sqrt3=-7\)
d: \(\frac{\sqrt{3-\sqrt5}\left(3+\sqrt5\right)}{\sqrt{10}+\sqrt2}\)
\(=\frac{\sqrt{6-2\sqrt5}\left(3+\sqrt5\right)}{2\left(\sqrt5+1\right)}\)
\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}\cdot\left(3+\sqrt5\right)}{2\left(\sqrt5+1\right)}=\frac{\left(\sqrt5-1\right)\left(3+\sqrt5\right)}{2\left(\sqrt5+1\right)}\)
\(=\frac{3\sqrt5+5-3-\sqrt5}{2\sqrt5+2}=\frac{2\sqrt5+2}{2\sqrt5+2}=1\)
