Lời giải:

\(\sqrt{\frac{(1+\sqrt{2})^3}{27}}=\sqrt{\frac{(1+\sqrt{2})^3}{3^3}}=\sqrt{\frac{3(1+\sqrt{2})^3}{3^4}}\)

\(=\frac{(1+\sqrt{2})\sqrt{3+3\sqrt{2}}}{9}\)

\(ab\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{(ab)^2(\frac{1}{a}+\frac{1}{b})}=\sqrt{ab^2+a^2b}\)

;

\(\sqrt{\dfrac{3}{\left(-4\right)^2}}=\dfrac{\sqrt{3}}{\sqrt{\left(-4\right)^2}}=\dfrac{\sqrt{3}}{4}\)

\(\sqrt{\dfrac{3}{\left(-4\right)^2}}=\dfrac{\sqrt{3}}{4}\)

1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)

\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)

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

\(=5\sqrt{3}\)

2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)

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

\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)

\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)

\(=\dfrac{-\sqrt{3}-5}{11}\)

3) Ta có: \(\sqrt{\dfrac{2}{5}}\)

\(=\dfrac{\sqrt{2}}{\sqrt{5}}\)

\(=\dfrac{\sqrt{2}\cdot\sqrt{5}}{5}\)

\(=\dfrac{\sqrt{10}}{5}\)

\(\sqrt{\frac{\left(1-\sqrt{3}\right)^2}{27}}=\sqrt{\frac{4-2\sqrt{3}}{27}}=0,1408832436\)

\(\sqrt{\dfrac{1}{600}}\)=\(\sqrt{\dfrac{1}{10^2\cdot6}}\)=\(\sqrt{\dfrac{1\cdot6}{10^2\cdot6\cdot6}}\)=\(\dfrac{\sqrt{6}}{60}\)

\(\sqrt{\dfrac{11}{540}}\)=\(\sqrt{\dfrac{11\cdot540}{540\cdot540}}\)=\(\dfrac{\sqrt{5940}}{540}\)=\(\dfrac{\sqrt{165}}{90}\)

\(\sqrt{\dfrac{3}{50}}\)=\(\sqrt{\dfrac{3\cdot50}{50\cdot50}}\)=\(\dfrac{\sqrt{150}}{50}\)=\(\dfrac{\sqrt{6}}{10}\)

\(\sqrt{\dfrac{5}{98}}\)=\(\sqrt{\dfrac{5\cdot98}{98\cdot98}}=\dfrac{\sqrt{490}}{98}=\dfrac{\sqrt{10}}{14}\)

\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)

\(\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt{6}}{60}\)

\(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{165}}{90}\)

\(\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt{6}}{10}\)

\(\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt{10}}{14}\)

\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)

EM thử thôi, ko chắc đâu ạ:( Sai thì xin thông cảm cho ạ.

1) \(\sqrt{\frac{2}{3-\sqrt{5}}}=\sqrt{\frac{2\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}=\sqrt{\frac{6+2\sqrt{5}}{4}}=\frac{\sqrt{6+2\sqrt{5}}}{2}\)

2) \(\sqrt{\frac{a-4}{2\left(\sqrt{a}-2\right)}}=\sqrt{\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{2\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}}\)

\(=\sqrt{\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{2\left(a-4\right)}}\)

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

4) \(\sqrt{\frac{a}{4-2\sqrt{3}}}=\sqrt{\frac{a\left(4+2\sqrt{3}\right)}{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}}=\sqrt{\frac{4a+2a\sqrt{3}}{16-12}}=\sqrt{\frac{4a+2a\sqrt{3}}{4}}=\frac{\sqrt{4a+2a\sqrt{3}}}{2}\)

a) \(\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt{1}}{10\sqrt{6}}=\dfrac{\sqrt{1}.\sqrt{6}}{10\sqrt{6}.\sqrt{6}}=\dfrac{\sqrt{6}}{60}\)

b) \(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{11}}{6\sqrt{15}}=\dfrac{\sqrt{11}.\sqrt{15}}{6\sqrt{15}.\sqrt{15}}=\dfrac{\sqrt{165}}{90}\)

c) \(\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt{3}}{5\sqrt{2}}=\dfrac{\sqrt{3}.\sqrt{2}}{5\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6}}{10}\)

d) \(\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt{5}}{7\sqrt{2}}=\dfrac{\sqrt{5}.\sqrt{2}}{7\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{10}}{14}\)

e) \(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{\sqrt{\left(1-\sqrt{3}\right)^2}}{3\sqrt{3}}=\dfrac{\sqrt{3}-1}{3\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{3\sqrt{3}.\sqrt{3}}=\dfrac{3-\sqrt{3}}{9}\)

\(\sqrt{\dfrac{1}{600}}=\sqrt{\dfrac{1\cdot6}{600\cdot6}}=\sqrt{\dfrac{6}{60^2}}=\dfrac{\sqrt{6}}{60}\)

\(\sqrt{\dfrac{11}{540}}=\sqrt{\dfrac{11\cdot15}{540\cdot15}}=\sqrt{\dfrac{165}{90^2}}=\dfrac{\sqrt{165}}{90}\)

\(\sqrt{\dfrac{3}{50}}=\sqrt{\dfrac{3\cdot2}{50\cdot2}}=\sqrt{\dfrac{6}{10^2}}=\dfrac{\sqrt{6}}{10}\)

\(\sqrt{\dfrac{5}{98}}=\sqrt{\dfrac{5\cdot2}{98\cdot2}}=\sqrt{\dfrac{10}{12^2}}=\dfrac{\sqrt{10}}{12}\)

\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\sqrt{\dfrac{3\left(1-\sqrt{3}\right)^2}{27\cdot3}}\)

\(=\dfrac{\sqrt{3\left(1-\sqrt{3}\right)^2}}{\sqrt{9^2}}=\dfrac{\left|1-\sqrt{3}\right|\cdot\sqrt{3}}{9}\)

\(=\dfrac{\left(\sqrt{3}-1\right)\sqrt{3}}{9}\)

a) \(\sqrt{\frac{3}{125}}=\frac{\sqrt{3.125}}{125}=\frac{\sqrt{375}}{125}=\frac{5\sqrt{15}}{125}=\frac{\sqrt{15}}{25}\)

b) \(\sqrt{\frac{3}{2a^3}}=\frac{\sqrt{3.2a^3}}{2a^3}=\frac{\sqrt{6a^3}}{2a^3}\)

c) \(\sqrt{\frac{\left(1-\sqrt{3}\right)^2}{27}}=\frac{\sqrt{27\left(1-\sqrt{3}\right)^2}}{27}=\frac{3.\left(\sqrt{3}-1\right)\sqrt{3}}{27}=\frac{\left(\sqrt{3}-1\right)\sqrt{3}}{9}\)

d) \(\sqrt{\frac{11}{540}}=\frac{\sqrt{11.540}}{540}=\frac{\sqrt{5940}}{50}=\frac{6\sqrt{165}}{50}=\frac{3\sqrt{165}}{25}\)