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.

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

a: \(\frac{1}{\sqrt5-\sqrt3-\sqrt2}=\frac{\sqrt5-\sqrt3+\sqrt2}{\left(\sqrt5-\sqrt3-\sqrt2\right)\left(\sqrt5-\sqrt3+\sqrt2\right)}\)

\(=\frac{\sqrt5-\sqrt3+\sqrt2}{8-2\sqrt{15}-2}=\frac{\sqrt5-\sqrt3+\sqrt2}{6-2\sqrt{15}}\)

\(=\frac{\left(\sqrt5-\sqrt3+\sqrt2\right)\left(3+\sqrt{15}\right)}{2\left(3-\sqrt{15}\right)\left(3+\sqrt{15}\right)}=\frac{\left(\sqrt5-\sqrt3+\sqrt2\right)\left(3+\sqrt{15}\right)}{2\left(9-15\right)}\)

\(=\frac{\left(\sqrt5-\sqrt3+\sqrt2\right)\left(3+\sqrt{15}\right)}{2\cdot\left(-6\right)}=\frac{\left(-\sqrt5+\sqrt3-\sqrt2\right)\left(3+\sqrt{15}\right)}{12}\)

b: \(\frac{2}{-1-\sqrt2+\sqrt3}=\frac{2}{\sqrt3-\sqrt2-1}=\frac{2\left(\sqrt3-\sqrt2+1\right)}{\left(\sqrt3-\sqrt2-1\right)\left(\sqrt3-\sqrt2+1\right)}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)

\(=\sqrt{x-1-2\sqrt{x-1+1}}+\sqrt{x-1+2\sqrt{x-1}+1}\)

\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)

\(=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)

\(=\sqrt{x-1}-1+\sqrt{x-1}+1\left(x\ge2\right)=2\sqrt{x-1}\)

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

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

\(=\dfrac{14\sqrt{5}+7\sqrt{3}}{17}\)

a: \(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\sqrt{a}\cdot\sqrt{a}-\sqrt{a}}{-\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{-\left(\sqrt{a}-1\right)}=-\sqrt{a}\)

b: \(\dfrac{2+\sqrt{3}}{2-\sqrt{7}}=\dfrac{\left(2+\sqrt{3}\right)\left(2+\sqrt{7}\right)}{4-7}\)

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

\(=\dfrac{-4-2\sqrt{7}-2\sqrt{3}-\sqrt{21}}{3}\)

c: \(3xy\cdot\sqrt{\dfrac{2}{xy}}=\dfrac{3xy}{\sqrt{xy}}\cdot\sqrt{2}=3\sqrt{2}\cdot\sqrt{xy}\)

d:

\(\dfrac{3}{\sqrt[3]{3}+\sqrt[3]{2}}=\dfrac{3\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}{3+2}\)

\(=\dfrac{3}{5}\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)\)

e:

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

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

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

\(=2\sqrt{3}+2-10-5\sqrt{3}-3-\sqrt{3}\)

\(=-4\sqrt{3}-11\)

f:

\(\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{9}}+\dfrac{1}{\sqrt{9}+\sqrt{13}}\)

\(=\dfrac{\sqrt{5}-1}{5-1}+\dfrac{\sqrt{9}-\sqrt{5}}{9-5}+\dfrac{\sqrt{13}-\sqrt{9}}{13-9}\)

\(=\dfrac{-1+\sqrt{5}-\sqrt{5}+\sqrt{9}-\sqrt{9}+\sqrt{13}}{4}=\dfrac{\sqrt{13}-1}{4}\)

\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\\ =\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{1-\sqrt{a}}\\ =\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\\ =-\sqrt{a}\\ \dfrac{2+\sqrt{3}}{2-\sqrt{7}}\\ =\dfrac{\left(2+\sqrt{3}\right)\left(2+\sqrt{7}\right)}{4-7}\\ =\dfrac{4+2\sqrt{7}+2\sqrt{3}+\sqrt{21}}{-3}\\\)

\(3xy\sqrt{\dfrac{2}{xy}}\\ =\sqrt{\dfrac{\left(3xy\right)^2\cdot2}{xy}}\\ =\sqrt{\dfrac{9x^2y^2\cdot2}{xy}}\\ =\sqrt{9xy\cdot2}\\ =\sqrt{18xy}\)

\(\dfrac{4}{\sqrt{3}+1}-\dfrac{5}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\\ =\dfrac{4\left(\sqrt{3}+1\right)}{3-1}-\dfrac{5\left(\sqrt{3}+2\right)}{3-4}+\dfrac{6\left(\sqrt{3}+3\right)}{3-9}\\ =\dfrac{4\left(\sqrt{3}+1\right)}{2}-\dfrac{5\left(\sqrt{3}+2\right)}{-1}+\dfrac{6\left(\sqrt{3}+3\right)}{-6}\\ =2\sqrt{3}+2+5\sqrt{3}+10-\sqrt{3}-3\\ =6\sqrt{3}+9\)

\(\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{9}}+\dfrac{1}{\sqrt{9}+\sqrt{13}}\\ =\dfrac{1-\sqrt{5}}{1-5}+\dfrac{\sqrt{5}-\sqrt{9}}{5-9}+\dfrac{\sqrt{9}-\sqrt{13}}{9-13}\\ =\dfrac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}}{-4}\\ =\dfrac{1-\sqrt{13}}{-4}\)

`# gvy`