\(A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}=\sqrt{\sqrt{5}^2+\sqrt{2}^2+1^2+2\sqrt{2}.1}+2\sqrt{5}.1+2\sqrt{2}\sqrt{5}}\)

=\(\sqrt{\left(\sqrt{5}+\sqrt{2}+1\right)^2=\sqrt{5}+\sqrt{2}+1}\)

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/ \(\sqrt{2}+\sqrt{6}\)

b/ Sửa đề:

\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=1\)

c/ \(1+\sqrt{2}+\sqrt{5}\)

giải rõ ra hộ mình với

a/ \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

\(=2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}\)

\(=2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)

\(=2\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}\)

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

\(=2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}\)

\(=2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=2\sqrt{3+\left(\sqrt{3}-1\right)}\)

\(=\sqrt{2}\sqrt{4+2\sqrt{3}}\)

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

\(=\sqrt{2}\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\sqrt{2}\left(\sqrt{3}+1\right)\)

\(=\sqrt{2}+\sqrt{6}\)

\(B=\sqrt{5}\left(\sqrt{20}-\sqrt{8}\right)+2\sqrt{10}\)

\(=\sqrt{100}-\sqrt{40}+2\sqrt{10}\)

\(=\sqrt{10^2}-\sqrt{2^2.10}+2\sqrt{10}\)

\(=10-2\sqrt{10}+2\sqrt{10}\)

\(=10+0\)

\(=10\)

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

`#Hưng`

\(a,3\sqrt{8\sqrt{5}}-2\sqrt{9\sqrt{20}}\\ =\sqrt{9.8\sqrt{5}}-\sqrt{4.9\sqrt{20}}\\ =\sqrt{72\sqrt{5}}-\sqrt{36\sqrt{20}}\\ =\sqrt{\sqrt{5184.5}}-\sqrt{\sqrt{1296.20}}\\ =\sqrt{\sqrt{25920}}-\sqrt{\sqrt{25920}}\\ =0\)

\(b,ĐKXĐ:x\sqrt{x}-\sqrt{x}+x-1\ne0\\ \Rightarrow\sqrt{x}\left(x-1\right)+\left(x-1\right)\ne0\\ \Rightarrow\left(x-1\right)\left(\sqrt{x}+1\right)\ne0\\ \Rightarrow x-1\ne0\left(vì.\sqrt{x}+1>0\right)\\ \Rightarrow x\ne1\)

a) \(A=\sqrt{18}.\sqrt{2}-\sqrt{48}:\sqrt{3}=\sqrt{18.2}-\sqrt{48:3}\)

\(=\sqrt{36}-\sqrt{16}=6-4=2\)

b) \(B=\dfrac{8}{\sqrt{5}-1}+\dfrac{8}{\sqrt{5}+1}=\dfrac{8\sqrt{5}+8+8\sqrt{5}-8}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\dfrac{16\sqrt{5}}{4}=4\sqrt{5}\)