a) `-2/(3\sqrt11) = (-2\sqrt11)/(3\sqrt11 .\sqrt11) =(-2\sqrt11)/(3.11)=(-2\sqrt11)/33`

b) `3/(\sqrt7+4) = (3.(\sqrt7-4))/((\sqrt7+4)(\sqrt7-4))`

`=(3.(\sqrt7-4))/((\sqrt7)^2-4^2)`

`=(3.(\sqrt7-4))/(-9)`

`=(4-\sqrt7)/3`

\(\dfrac{-2}{3\sqrt{11}}=\dfrac{-2\sqrt{11}}{33}\)

\(\dfrac{3}{4+\sqrt{7}}=\dfrac{12-3\sqrt{7}}{7}\)

Ta có: \(\frac{10}{\sqrt{3}-1}\)

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

\(=5\sqrt{3}+5\)

\(=\frac{6\sqrt{2}\left(3\sqrt{7}-5\sqrt{2}\right)}{2\left(3\sqrt{7}-5\sqrt{2}\right)}=\frac{6\sqrt{2}}{2}=3\sqrt{2}\)

\(\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}=\dfrac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\dfrac{2-\sqrt{3}}{4-3}+\dfrac{2+\sqrt{3}}{4-3}=2-\sqrt{3}+2+\sqrt{3}=4\)

1)

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

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

\(\dfrac{5}{\sqrt{7}}=\dfrac{5\sqrt{7}}{\sqrt{49}}=\left(\dfrac{5}{7}\right)\sqrt{7}\)

1:

\(\dfrac{2\sqrt{3}}{5\sqrt{7}}=\dfrac{2\sqrt{21}}{35}\)

\(\dfrac{5}{2\sqrt{3}}=\dfrac{5\sqrt{3}}{6}\)

2: \(\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\)

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

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

\(\dfrac{12}{\sqrt{5}-\sqrt{3}}=6\left(\sqrt{5}+\sqrt{3}\right)=6\sqrt{5}+6\sqrt{3}\)

1) \(5\sqrt{8}-\dfrac{7}{2}\sqrt{72}+6\sqrt{\dfrac{1}{2}}\\ =5.\sqrt{4^2.\dfrac{1}{2}}-\dfrac{7}{2}.\sqrt{12^2.\dfrac{1}{2}}+6.\sqrt{\dfrac{1}{2}}=\left(5.4+\dfrac{7}{2}.12+6\right)\sqrt{\dfrac{1}{2}}\\ =68\sqrt{\dfrac{1}{2}}\)

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

\(=\dfrac{1-\sqrt[3]{2}+\sqrt[3]{4}}{1-2}=\dfrac{1-\sqrt[3]{2}+\sqrt[3]{4}}{-1}=\sqrt[3]{2}-\sqrt[3]{4}-1\)