a: \(\dfrac{6}{5\sqrt{8}}=\dfrac{6}{10\sqrt{2}}=\dfrac{3}{5\sqrt{2}}=\dfrac{3\sqrt{2}}{10}\)

b: \(\dfrac{7}{5+2\sqrt{3}}=\dfrac{7\left(5-2\sqrt{3}\right)}{13}\)

c: \(\dfrac{6}{\sqrt{7}-\sqrt{5}}=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}=3\left(\sqrt{7}+\sqrt{5}\right)\)

a) \(\dfrac{6}{5\sqrt{8}}\)

\(=\dfrac{6}{5\cdot2\sqrt{2}}\)

\(=\dfrac{6}{10\sqrt{2}}\)

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

\(=\dfrac{3\sqrt{2}}{10}\)

b) \(\dfrac{7}{5+2\sqrt{3}}\)

\(=\dfrac{7\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}\)

\(=\dfrac{7\left(5-2\sqrt{3}\right)}{5^2-\left(2\sqrt{3}\right)^2}\)

\(=\dfrac{7\left(5-2\sqrt{3}\right)}{13}\)

\(=\dfrac{35-14\sqrt{3}}{13}\)

c) \(\dfrac{6}{\sqrt{7}-\sqrt{5}}\)

\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)

\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}\)

\(=3\sqrt{7}+3\sqrt{5}\)

\(\dfrac{4}{\sqrt{5}-\sqrt{2}}+\dfrac{3}{\sqrt{5}-2}-\dfrac{2}{\sqrt{3}-2}-\dfrac{\sqrt{3}-1}{6}\)

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

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

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

\(=\dfrac{8\left(\sqrt{2}+\sqrt{5}\right)}{6}+\dfrac{18\left(\sqrt{5}+2\right)}{6}+\dfrac{12\left(\sqrt{3}+2\right)}{6}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{8\sqrt{2}+8\sqrt{5}+18\sqrt{5}+36+12\sqrt{3}+24-\sqrt{3}+1}{6}\)

\(=\dfrac{8\sqrt{2}+26\sqrt{5}+11\sqrt{3}+61}{6}\)

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

\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)

\(=\dfrac{26\sqrt{5}+4\sqrt{2}+36+25+11\sqrt{3}}{6}\)

\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+4\sqrt{2}}{6}\)

Mình sửa lại chút nha bạn

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

\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)

\(=\dfrac{26\sqrt{5}+8\sqrt{2}+36+25+11\sqrt{3}}{6}\)

\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+8\sqrt{2}}{6}\)

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`

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

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

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

\(A=\frac{12}{\sqrt{7+2\sqrt{6}}}=\frac{12}{\sqrt{\left(\sqrt{6}+1\right)^2}}=\frac{12}{\sqrt{6}+1}=\frac{12\left(\sqrt{6}-1\right)}{5}\)

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

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

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

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

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