\(\dfrac{-2}{\sqrt{3}-1}=\dfrac{-2\left(\sqrt{3}+1\right)}{2}=-\sqrt{3}-1\\ \dfrac{\sqrt{5}}{\sqrt{7}-3}=\dfrac{-\sqrt{5}\left(\sqrt{7}+3\right)}{2}\\ \dfrac{3\sqrt{3}-2}{1-2\sqrt{3}}=\dfrac{\left(3\sqrt{3}-2\right)\left(1+2\sqrt{3}\right)}{-11}=\dfrac{\sqrt{3}-16}{11}\\ \dfrac{14}{\sqrt{10}+\sqrt{3}}=\dfrac{14\left(\sqrt{10}-\sqrt{3}\right)}{7}=2\sqrt{10}-2\sqrt{3}\)

Uii em cảm ơn ạ:3

1)

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

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

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

lười :v

a) \(\dfrac{14}{2\sqrt{3}-\sqrt{5}}\)

\(=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{\left(2\sqrt{3}-\sqrt{5}\right)\left(2\sqrt{3}+\sqrt{5}\right)}\)

\(=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{\left(2\sqrt{3}\right)^2-\sqrt{5^2}}=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{12-5}\)

\(=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{7}=2\left(2\sqrt{3}+\sqrt{5}\right)\)

\(=4\sqrt{3}+2\sqrt{5}\)

b) \(\dfrac{x^2-y}{x-\sqrt{y}}=\dfrac{\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)}{x-\sqrt{y}}=x+\sqrt{y}\)

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) \(\dfrac{6}{\sqrt{7}-\sqrt{5}}\)

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

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

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

a: \(\frac{7}{\sqrt5-\sqrt3-\sqrt7}=\frac{7\left(\sqrt5-\sqrt3+\sqrt7\right)}{\left(\sqrt5-\sqrt3\right)^2-7}\)

\(=\frac{7\left(\sqrt5-\sqrt3+\sqrt7\right)}{8-2\sqrt{15}-7}=\frac{7\left(\sqrt5-\sqrt3+\sqrt7\right)}{1-2\sqrt{15}}\)

\(=\frac{7\left(\sqrt5+\sqrt7-\sqrt3\right)\left(1+2\sqrt{15}\right)}{\left(1-2\sqrt{15}\right)\left(1+2\sqrt{15}\right)}=\frac{7\left(\sqrt5+\sqrt7-\sqrt3\right)\left(2\sqrt{15}+1\right)}{1-60}=\frac{-7\left(\sqrt5+\sqrt7-\sqrt3\right)\left(2\sqrt{15}+1\right)}{59}\)

b: \(\frac{5}{2-\sqrt3-\sqrt5}=\frac{5\left(2-\sqrt3+\sqrt5\right)}{\left(2-\sqrt3\right)^2-5}\)

\(=\frac{5\left(2-\sqrt3+\sqrt5\right)}{7-4\sqrt3-5}=\frac{5\left(2-\sqrt3+\sqrt5\right)}{-2\sqrt3+2}\)

\(=\frac{5\left(2-\sqrt3+\sqrt5\right)\left(2+2\sqrt3\right)}{\left(2-2\sqrt3\right)\left(2+2\sqrt3\right)}=\frac{5\left(2-\sqrt3+\sqrt5\right)\left(2+2\sqrt3\right)}{4-12}\)

\(=\frac{5\left(2-\sqrt3+\sqrt5\right)\left(2+2\sqrt3\right)}{-8}=\frac{-5\left(2-\sqrt3+\sqrt5\right)\left(\sqrt3+1\right)}{4}\)

a) \(\dfrac{7}{\sqrt{5}-\sqrt{3}-\sqrt{7}}\)

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

\(=\dfrac{7\sqrt{5}-7\sqrt{3}+7\sqrt{7}}{8-2\sqrt{15}-7}\)

\(=\dfrac{7\sqrt{5}-7\sqrt{3}+7\sqrt{7}}{1-2\sqrt{15}}\)

\(=\dfrac{\left(7\sqrt{5}-7\sqrt{3}+7\sqrt{7}\right)\left(1+2\sqrt{15}\right)}{1-60}\)

\(=\dfrac{7\sqrt{5}+70\sqrt{3}-7\sqrt{3}-42\sqrt{5}+7\sqrt{7}+14\sqrt{105}}{-59}\)

\(=\dfrac{-35\sqrt{5}+63\sqrt{3}+7\sqrt{7}+14\sqrt{105}}{-59}\)

\(=\dfrac{35\sqrt{5}-63\sqrt{3}-7\sqrt{7}-14\sqrt{105}}{59}\)

a: \(\frac{7}{\sqrt5-\sqrt3-\sqrt7}=\frac{7\left(\sqrt5-\sqrt3+\sqrt7\right)}{\left(\sqrt5-\sqrt3\right)^2-7}\)

\(=\frac{7\left(\sqrt5-\sqrt3+\sqrt7\right)}{8-2\sqrt{15}-7}=\frac{7\left(\sqrt5-\sqrt3+\sqrt7\right)}{1-2\sqrt{15}}\)

\(=\frac{7\left(\sqrt5+\sqrt7-\sqrt3\right)\left(1+2\sqrt{15}\right)}{\left(1-2\sqrt{15}\right)\left(1+2\sqrt{15}\right)}=\frac{7\left(\sqrt5+\sqrt7-\sqrt3\right)\left(2\sqrt{15}+1\right)}{1-60}=\frac{-7\left(\sqrt5+\sqrt7-\sqrt3\right)\left(2\sqrt{15}+1\right)}{59}\)

b: \(\frac{5}{1-\sqrt2-\sqrt3}\)

\(=\frac{-5}{\sqrt3+\sqrt2-1}=\frac{-5\left(\sqrt3+\sqrt2+1\right)}{\left(\sqrt3+\sqrt2-1\right)\left(\sqrt3+\sqrt2+1\right)}\)

\(=\frac{-5\left(\sqrt3+\sqrt2+1\right)}{\left(\sqrt3+\sqrt2\right)^2-1}=\frac{-5\left(\sqrt3+\sqrt2+1\right)}{4+2\sqrt6}\)

\(=\frac{-5\left(\sqrt3+\sqrt2+1\right)\left(\sqrt6-2\right)}{2\left(\sqrt6+2\right)\left(\sqrt6-2\right)}=\frac{-5\left(\sqrt3+\sqrt2+1\right)\left(\sqrt6-2\right)}{2\left(6-4\right)}=\frac{-5\left(\sqrt3+\sqrt2+1\right)\left(\sqrt6-2\right)}{4}\)