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

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

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

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

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

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

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

=0

a/ \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}=2\sqrt{4.2.5\sqrt{4.3}}-2\sqrt{\sqrt{25.3}}-3\sqrt{5\sqrt{16.3}}\)

= \(2.2\sqrt{2.5.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}=4.2\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-3.2\sqrt{5\sqrt{3}}\)

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

b/ \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}=2\sqrt{2.4\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{4.5\sqrt{3}}\)

= \(4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)

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

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

  \(=0\)

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

\(I=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)

\(=\left(2\sqrt{3}-5\sqrt{3}.\sqrt{3^2}+2\sqrt{2^2}.\sqrt{3}\right):\sqrt{3}\)

\(=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)

\(=-5\sqrt{3}.\dfrac{1}{\sqrt{3}}\)

\(=-5\)

\(K=\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\)

\(=\sqrt{5^2.5}-4\sqrt{3^2.5}+3\sqrt{2^2.5}-\sqrt{4^2.5}\)

\(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\)

\(=\sqrt{5}.\left(5-12+6-4\right)\)

\(=-5\sqrt{5}\)

\(L=2\sqrt{9}+\sqrt{25}-5\sqrt{4}\)

\(=2\sqrt{3^2}+\sqrt{5^2}-5\sqrt{2^2}\)

\(=2.3+5-5.2\)

\(=1\)

\(N=2\sqrt{32}-5\sqrt{27}-4\sqrt{8}+3\sqrt{75}\)

\(=2.4\sqrt{2}-5.3\sqrt{3}-4.2\sqrt{2}+3.5\sqrt{3}\)

\(=8\sqrt{2}-8\sqrt{2}-15\sqrt{3}+15\sqrt{3}\)

\(=0\)

\(O=2\sqrt{3.5^2}-3\sqrt{3.2^2}+\sqrt{3.3^2}\)

\(=2.5\sqrt{3}-3.2\sqrt{3}+3\sqrt{3}\)

\(=10\sqrt{3}-6\sqrt{3}+3\sqrt{3}\)

\(=7\sqrt{3}\)

\(L=\dfrac{2\sqrt{3}-15\sqrt{3}+8\sqrt{3}}{\sqrt{3}}=2-15+8=-5\)

\(K=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)

L=2*3+5-5*2=5-4=1

N=8căn 2-8căn2-15căn3+15căn 3=0

O=10căn 3-6căn3+3căn3=7căn 3

a: \(A=6\sqrt{3}+10\sqrt{3}-12\sqrt{3}=4\sqrt{3}\)

b: \(B=7\sqrt{3}+5\sqrt{3}-12\sqrt{3}=0\)

c: \(=12\sqrt{2}-6+3\left(9-4\sqrt{2}\right)=12\sqrt{2}-6+27-12\sqrt{2}=21\)

d: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)