a: Ta có: \(\dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}\left(\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{\sqrt{14}}{7}\)
b: Ta có: \(\left(\sqrt{12}+3\sqrt{15}-4\sqrt{135}\right)\cdot\sqrt{3}\)
\(=\left(\sqrt{12}+3\sqrt{15}-12\sqrt{15}\right)\cdot\sqrt{3}\)
\(=\left(2\sqrt{3}-9\sqrt{15}\right)\cdot\sqrt{3}\)
\(=6-27\sqrt{5}\)
d: Ta có: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{3}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1-\sqrt{3}\)
c: Ta có: \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\cdot\sqrt{80\sqrt{3}}-2\cdot\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=2\cdot4\sqrt{5}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-3\cdot2\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=0\)
