a) Ta có: \(\dfrac{1}{\sqrt{5}-\sqrt{3}-\sqrt{2}}\)

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

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

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

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

b) Ta có: \(\dfrac{2}{-1-\sqrt{2}+\sqrt{3}}\)

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

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

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

\(a,\dfrac{7}{\sqrt{12}}=\dfrac{7\sqrt{3}}{\sqrt{12}\cdot\sqrt{3}}\)

\(=\dfrac{7\sqrt{3}}{\sqrt{36}}=\dfrac{7\sqrt{3}}{6}\)

\(b,\dfrac{3}{2\sqrt{3}}=\dfrac{3\sqrt{3}}{2\sqrt{3}\cdot\sqrt{3}}\)

\(=\dfrac{3\sqrt{3}}{2\cdot3}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)

\(c,\dfrac{1}{5\sqrt{12}}=\dfrac{\sqrt{3}}{5\cdot2\sqrt{3}\cdot\sqrt{3}}\)

\(=\dfrac{\sqrt{3}}{10\cdot3}=\dfrac{\sqrt{3}}{30}\)

\(d,\dfrac{2\sqrt{3}+3}{4\sqrt{3}}=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{4\sqrt{3}}\)

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

a) \(\dfrac{7}{\sqrt[]{12}}=\dfrac{7}{2\sqrt[]{3}}=\dfrac{7\sqrt[]{3}}{2\sqrt[]{3}.\sqrt[]{3}}=\dfrac{7\sqrt[]{3}}{6}\)

b) \(\dfrac{3}{2\sqrt[]{3}}=\dfrac{\sqrt[]{3}.\sqrt[]{3}}{2\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{2}\)

c) \(\dfrac{1}{5\sqrt[]{12}}=\dfrac{1}{10\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{10\sqrt[]{3}.\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{30}\)

d) \(\dfrac{2\sqrt[]{3}+3}{4\sqrt[]{3}}=\dfrac{\sqrt[]{3}\left(2\sqrt[]{3}+3\right)}{4\sqrt[]{3}.\sqrt[]{3}}=\dfrac{3\left(2+\sqrt[]{3}\right)}{12}=\dfrac{2+\sqrt[]{3}}{4}\)

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) Ta có: \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{7}}\)

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

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

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

\(=\dfrac{-7\left(\sqrt{5}+10\sqrt{3}-\sqrt{3}-6\sqrt{5}-\sqrt{7}-2\sqrt{105}\right)}{59}\)

\(=\dfrac{-7\left(-5\sqrt{5}+9\sqrt{3}-\sqrt{7}-2\sqrt{105}\right)}{59}\)

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

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

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

\(=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{9-5}}\)

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

b) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)

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

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

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

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

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

a: \(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}}=\dfrac{3-\sqrt{5}}{2}\)

b: \(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{1}}=2-\sqrt{3}\)

d: \(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

=(căn 6-11)(căn 6+11)

=6-121=-115

Lời giải:

a.

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

b.

\(=\frac{2[(\sqrt{3}-(\sqrt{2}-1)]}{[(\sqrt{3}+(\sqrt{2}-1)][\sqrt{3}-(\sqrt{2}-1)]}=\frac{2(\sqrt{3}-\sqrt{2}+1)}{3-(\sqrt{2}-1)^2}=\frac{2(\sqrt{3}-\sqrt{2}+1)}{2\sqrt{2}}\)

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

c.

\(=\frac{5(\sqrt[3]{2^2}-3\sqrt[3]{2}+3^2)}{(\sqrt[3]{2})^3+3^3}=\frac{5(\sqrt[3]{4}+3\sqrt[3]{2}+9)}{29}\)