d: Ta có: \(\dfrac{3\sqrt{15}-\sqrt{5}}{3\sqrt{3}-1}-\dfrac{2}{\sqrt{5}+\sqrt{3}}+\sqrt{12-6\sqrt{3}}\)
\(=\sqrt{5}-\sqrt{5}+\sqrt{3}+3-\sqrt{3}\)
=3
\(c,=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{3\left(\sqrt{7}-2\right)}{7-4}+\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}\\ =\sqrt{5}-\sqrt{7}+2+\sqrt{5}+\sqrt{7}\\ =2\sqrt{5}+2\\ d,=\dfrac{\sqrt{5}\left(3\sqrt{3}-1\right)}{3\sqrt{3}-1}-\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}+\sqrt{\left(3-\sqrt{3}\right)^2}\\ =\sqrt{5}-\sqrt{5}+\sqrt{3}+3-\sqrt{3}=3\\ e,=\sqrt{3}-\sqrt{6+6\sqrt{3\sqrt{3}+4-\sqrt{3}}}\\ =\sqrt{3}-\sqrt{6+6\sqrt{4+2\sqrt{3}}}\\ =\sqrt{3}-\sqrt{6+6\left(\sqrt{3}+1\right)}\\ =\sqrt{3}-\sqrt{12+6\sqrt{3}}=\sqrt{3}-\left(3+\sqrt{3}\right)=-3\)
c) \(=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{3\left(\sqrt{7}-2\right)}{7-4}+\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}\)
\(=\sqrt{5}-\sqrt{7}+2+\sqrt{7}+\sqrt{5}=2+2\sqrt{5}\)
d) \(=\dfrac{\sqrt{5}\left(3\sqrt{3}-1\right)}{3\sqrt{3}-1}-\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}+\sqrt{\left(3-\sqrt{3}\right)^2}\)
\(=\sqrt{5}-\sqrt{5}+\sqrt{3}+3-\sqrt{3}=3\)
e) \(=\sqrt{3}-\sqrt{6+6\sqrt{3\sqrt[]{3}+\sqrt{\left(4-\sqrt{3}\right)^2}}}\)
\(=\sqrt{3}-\sqrt{6+6\sqrt{3\sqrt{3}+4-\sqrt{3}}}\)
\(=\sqrt{3}-\sqrt{6+6\sqrt{4+2\sqrt{3}}}\)
\(=\sqrt{3}-\sqrt{6+6\sqrt{\left(\sqrt{3}+1\right)^2}}=\sqrt{3}-\sqrt{6+6\left(\sqrt{3}+1\right)}\)
\(=\sqrt{3}-\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{3}-\sqrt{\left(3+\sqrt{3}\right)^2}=\sqrt{3}-3-\sqrt{3}=-3\)
