a) \(5\sqrt{12}+4\sqrt{27}-6\sqrt{48}=10\sqrt{3}+12\sqrt{3}-24\sqrt{3}=-2\sqrt{3}\)
b) \(\left(\sqrt{300}-2\sqrt{675}+5\sqrt{75}\right):\sqrt{3}=\left(10\sqrt{3}-30\sqrt{3}+25\sqrt{3}\right):\sqrt{3}\)
\(=10-30+25=5\)
d) \(\dfrac{2}{4-2\sqrt{3}}+\dfrac{2}{4+2\sqrt{3}}=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
\(=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=4\)
a: Ta có: \(5\sqrt{12}+4\sqrt{27}-6\sqrt{48}\)
\(=10\sqrt{3}+12\sqrt{3}-24\sqrt{3}\)
\(=-2\sqrt{3}\)
b: Ta có: \(\left(\sqrt{300}-2\sqrt{675}+5\sqrt{75}\right):\sqrt{3}\)
\(=\left(10\sqrt{3}-30\sqrt{3}+25\sqrt{3}\right):\sqrt{3}\)
=5
