\(=2\sqrt{80\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-4\sqrt{45\sqrt{3}}\)

\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-12\sqrt{5\sqrt{3}}\)

=0

a: \(=2\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)

\(=4\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}\)

b: \(=2\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}\)

a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)

  \(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)

  \(=0\)

b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)

Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)

\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)

Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)

a) E=0

b) F=căn 3/2

a) \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

\(=2\sqrt{40.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}\)

\(=\left(2\sqrt{80}-2\sqrt{5}-3\sqrt{20}\right).\sqrt{\sqrt{3}}\)

\(=\left(8\sqrt{5}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}=0\)

b) \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

\(=\left(4\sqrt{2}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}\)

\(=\left(4\sqrt{2}-8\sqrt{5}\right).\sqrt{\sqrt{3}}\)

\(=\sqrt{\sqrt{3}}\left(\sqrt{2}-2\sqrt{5}\right)\)

a)2\(\sqrt{40\sqrt{12}}\) -2\(\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

=\(2\sqrt{40.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

=\(2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

=\(8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)

=0

b)\(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

=\(4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)

=4\(\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)

\(A=\sqrt{12}+2\sqrt{27}+3\sqrt{45}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+9\sqrt{5}-36\sqrt{3}\)

\(=9\sqrt{5}-28\sqrt{3}\)

\(B=\left(\sqrt{48}-2\sqrt{75}+\sqrt{108}-\sqrt{147}\right):\sqrt{3}\)

\(=4-2\cdot5+6-7\)

\(=4-10+6-7\)

=-7

A=\(\sqrt{12}\)+2\(\sqrt{27}\)+3\(\sqrt{45}\) -9\(\sqrt{48}\)

  =\(\sqrt{4.3}\) +2\(\sqrt{9.3}\)+3\(\sqrt{9.5}\) -9\(\sqrt{16.3}\)

  =2\(\sqrt{3}\) +6\(\sqrt{3}\)+9\(\sqrt{5}\) -36\(\sqrt{3}\)

   =\(\sqrt{3}\)(2+6-36) + 9\(\sqrt{5}\)

   =9\(\sqrt{5}\)- 28\(\sqrt{3}\)