\(A=2\cdot\sqrt{9+4\sqrt{5}}+\sqrt{5}-3\sqrt{5}\)

=2(căn 5+2)-2căn 5

=4

1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)

\(=\dfrac{20-6}{2}=7\)

\(A=\dfrac{\left(1+\sqrt{5}\right)^2-\left(1-\sqrt{5}\right)^2}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}=\dfrac{6+2\sqrt{5}-6+2\sqrt{5}}{1-5}=\dfrac{4\sqrt{5}}{-4}=-\sqrt{5}\)

\(A=\dfrac{\sqrt{6+2\sqrt{5}}}{2-\sqrt{6-2\sqrt{5}}}-\dfrac{\sqrt{6-2\sqrt{5}}}{2+\sqrt{6+2\sqrt{5}}}\)

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

\(=\dfrac{\left(3+\sqrt{5}\right)\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}{4}\)

\(=\dfrac{3\sqrt{5}+3+5+\sqrt{5}-3\sqrt{5}+5+3-\sqrt{5}}{4}\)

\(=4\)

a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}\)

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

Câu a bạn xem lại đề nhé vì \(\sqrt{2-6}=\sqrt{-4}\left(VLý\right)\)

b) \(\dfrac{3\sqrt{5}+5\sqrt{3}}{\sqrt{3}+\sqrt{5}}=\dfrac{\sqrt{3.5}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}}=\sqrt{15}\)

c) \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)

a) \(\dfrac{3\sqrt{2}-6}{\sqrt{2}-1}\)

\(a,=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)^2}+\left|2-\sqrt{5}\right|=3-\sqrt{5}+\sqrt{5}-2=1\\ c,=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}-\dfrac{-\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=\sqrt{5}-\sqrt{3}+\sqrt{3}=\sqrt{5}\)

h) Ta có: \(\sqrt{\dfrac{3+\sqrt{5}}{\sqrt{3-\sqrt{5}}}}+\sqrt{\dfrac{3-\sqrt{5}}{\sqrt{3+\sqrt{5}}}}\)

\(=\sqrt{\dfrac{6+2\sqrt{5}}{\sqrt{2}\left(\sqrt{5}-1\right)}}+\sqrt{\dfrac{6-2\sqrt{5}}{\sqrt{2}\left(\sqrt{5}+1\right)}}\)

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

\(=\dfrac{4\sqrt{2}}{\sqrt{2}\left(\sqrt{5}-1\right)}+\dfrac{4\sqrt{2}}{\sqrt{2}\left(\sqrt{5}+1\right)}\)

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

a: \(E=1+1=2\)

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

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

d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)