Trả lời:

\(A=\sqrt{3}-\frac{\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)

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

\(A=\sqrt{3}+\frac{\sqrt{6}.\left(\sqrt{2}+1\right)}{2-1}-\frac{2.\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)

\(A=\sqrt{3}+\sqrt{6}.\left(\sqrt{2}+1\right)-2\)

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

\(A=\sqrt{3}+2\sqrt{3}+\sqrt{6}-2\)

\(A=3\sqrt{3}+\sqrt{6}-2\)

Mình nhầm 

a) \(\left(\sqrt{\dfrac{9}{20}}-\sqrt{\dfrac{1}{2}}\right).\sqrt{2}=\sqrt{\dfrac{9}{20}.2}-\sqrt{\dfrac{1}{2}.2}=\sqrt{\dfrac{9}{10}}-1=\dfrac{3}{\sqrt{10}}-1\)

\(=\dfrac{3\sqrt{10}}{10}-1\)

b) \(\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right)\sqrt{3}=\sqrt{12.3}+\sqrt{27.3}-\sqrt{3.3}\)

\(=\sqrt{36}+\sqrt{81}-\sqrt{9}=6+9-3=12\)

c) \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right)\sqrt{6}=\sqrt{\dfrac{8}{3}.6}-\sqrt{24.6}+\sqrt{\dfrac{50}{3}.6}\)

\(=\sqrt{16}-\sqrt{144}+\sqrt{100}=4-12+10=2\)

a: \(\left(2\sqrt6-4\sqrt3+5\sqrt2-\frac14\cdot\sqrt8\right)\cdot3\sqrt6\)

\(=\left(2\sqrt6-4\sqrt3+5\sqrt2-\frac12\sqrt2\right)\cdot3\sqrt6\)

\(=\left(2\sqrt6-4\sqrt3+\frac92\cdot\sqrt2\right)\cdot3\sqrt6\)

\(=2\sqrt6\cdot3\sqrt6-4\sqrt3\cdot3\sqrt6+\frac92\cdot\sqrt2\cdot3\sqrt6\)

\(=36-12\sqrt{18}+\frac{27}{2}\sqrt{12}=36-36\sqrt2+27\sqrt3\)

b: \(\left(\sqrt{\frac17}-\sqrt{\frac{16}{7}}+\sqrt7\right):\sqrt7=\left(\frac{\sqrt7}{7}-\frac{4\sqrt7}{7}+\sqrt7\right):\sqrt7\)

\(=\frac17-\frac47+1=\frac87-\frac47=\frac47\)

c: \(\left(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\right)^2\)

\(=3-\sqrt5+3+\sqrt5+2\cdot\sqrt{\left(3-\sqrt5\right)\left(3+\sqrt5\right)}\)

\(=6+2\cdot\sqrt{9-5}=6+2\cdot2=10\)

bạn viết công thức đi

viết này ai hiểu

\(3-\sqrt{3}+\sqrt{6}=\left(\sqrt{3}\right)^2-\sqrt{3}+\sqrt{3}.\sqrt{2}\)

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

à cộng \(\sqrt{2}\)bạn nhé

a: Ta có: \(\dfrac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}\)

\(=\dfrac{\sqrt{10}\left(2+\sqrt{3}\right)-\sqrt{2}\left(2+\sqrt{3}\right)}{2\sqrt{2}\left(\sqrt{5}-1\right)}\)

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

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

b) Ta có: \(\sqrt{\left(1-\sqrt{2006}\right)^2}\cdot\sqrt{2007+2\sqrt{2006}}\)

\(=\left(\sqrt{2006}-1\right)\left(\sqrt{2006}+1\right)\)

=2005