a)\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}=5\sqrt{3}-\frac{\sqrt{15}}{3}+3\sqrt{3}+6\sqrt{3}=14\sqrt{3}-\frac{\sqrt{15}}{3}\)

b) \(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}=4\sqrt{3}+\frac{\sqrt{15}}{3}+10\sqrt{3}-\frac{5\sqrt{3}}{3}=\frac{12\sqrt{3}+30\sqrt{3}-5\sqrt{3}}{3}+\frac{\sqrt{15}}{3}=\frac{37\sqrt{3}+\sqrt{15}}{3}\)

c) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=\left[\left(\sqrt{15}\right)^2+4\sqrt{45}+\left(2\sqrt{3}\right)^2\right]+12\sqrt{5}=15+12\sqrt{5}+12+12\sqrt{5}=27+24\sqrt{5}\)

d) \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{18}-\sqrt{12}+\sqrt{6}-2\sqrt{2}=3\sqrt{2}-2\sqrt{3}+\sqrt{6}-2\sqrt{2}=\sqrt{2}-2\sqrt{3}+\sqrt{6}\)

e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=\left(\sqrt{3}\right)^2+2\sqrt{3}+1-2\sqrt{3}+4=3+2\sqrt{3}+1-2\sqrt{3}+4=8\)

f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)

g) \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}=\left(\frac{\sqrt{5}+2-\sqrt{5}+2+5-2}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\right)\frac{1}{3+2\sqrt{2}}=\frac{7}{3}.\frac{1}{3+2\sqrt{2}}=\frac{7}{9+6\sqrt{2}}\)

a) \(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)

\(=\sqrt{75}-\sqrt{\frac{16}{3}}+\frac{9}{2}\sqrt{\frac{8}{3}}+2\sqrt{27}\)

\(=5\sqrt{3}-\frac{4}{\sqrt{3}}+3\sqrt{6}+6\sqrt{3}\)

\(=-\frac{4}{\sqrt{3}}+5\sqrt{3}+3\sqrt{6}+6\sqrt{3}\)

\(=-\frac{4}{\sqrt{3}}+11\sqrt{3}+3\sqrt{6}\)

\(=-\frac{4\sqrt{3}}{3}+11\sqrt{3}+3\sqrt{6}\)

b) \(\sqrt{48}-\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)

\(=\sqrt{48}-\sqrt{\frac{16}{3}}+2\sqrt{75}-5\sqrt{\frac{4}{3}}\)

\(=4\sqrt{3}-\frac{4}{\sqrt{3}}+10\sqrt{3}-\frac{10}{\sqrt{3}}\)

\(=-\frac{4}{\sqrt{3}}-\frac{10}{\sqrt{3}}+4\sqrt{3}+10\sqrt{3}\)

\(=-\frac{14\sqrt{3}}{3}+4\sqrt{3}+10\sqrt{3}\)

\(=-\frac{14\sqrt{3}}{3}+14\sqrt{3}\)

c)\(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)

\(=27+12\sqrt{5}+12\sqrt{5}\)

\(=27+24\sqrt{5}\)

d)\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)

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

e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)

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

= 8

f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)

\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

\(=\frac{14}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

= 14

a) \(2\sqrt{2}.\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)

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

\(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}\)

= 9 (đpcm)

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

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

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

a, = \(\frac{\sqrt{15}}{10}\) + \(\frac{\sqrt{15}}{30}\) - \(\frac{2\sqrt{15}}{15}\)

= \(\sqrt{15}\left(\frac{1}{10}+\frac{1}{30}-\frac{2}{15}\right)\)

= \(\sqrt{15}.0\)

= 0

b, = \(\left(\frac{\sqrt{5}+\sqrt{3}}{5-3}+\frac{\sqrt{5}-\sqrt{3}}{5-3}\right).\sqrt{5}\)

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

= \(\frac{2\sqrt{5}}{2}.\sqrt{5}\)

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

= 5

c, = \(\frac{5\sqrt{3}}{\sqrt{15}}+\frac{3\sqrt{5}}{\sqrt{15}}\)

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

d, Mình sửa lại đề bài cho bạn : \(\left(2+\sqrt{5}\right)^2-\left(2-\sqrt{5}\right)^2\)

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

= \(2\sqrt{5}.4\)

= \(8\sqrt{5}\)

e, = \(\frac{4\sqrt{3}}{3}+15\sqrt{3}-3\sqrt{3}-\frac{20\sqrt{3}}{3}\)

= \(\sqrt{3}.\left(\frac{4}{3}+15-3-\frac{20}{3}\right)\)

= \(\sqrt{3}.\frac{20}{3}\)

= \(\frac{20\sqrt{3}}{3}\)

a, $\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}$

b, $\left(\frac{1}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{3}}\right).\sqrt{5}$

c, $\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}$

d, ${\left(2+\sqrt{5}\right)}^2-{\left(2+\sqrt{5}\right)}^2$

e, $\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}$

a) Ta có: \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)

\(=\sqrt{3}\left(2+\sqrt{16}-\sqrt{25}-\sqrt{81}\right)\)

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

\(=-8\sqrt{3}\)

b) Ta có: \(\left(\frac{\sqrt{7}-\sqrt{14}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}+\sqrt{5}}\)

\(=\left(\frac{\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\cdot\left(\sqrt{7}+\sqrt{5}\right)\)

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

\(=7-5=2\)

c) Ta có: \(\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)

\(=\left(\sqrt{3}+1\right)\cdot\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)

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

\(=\left(\sqrt{3}+1\right)\cdot\left|\sqrt{3}-1\right|\)

\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)(Vì \(\sqrt{3}>1\))

\(=3-1=2\)

d) Ta có: \(5\sqrt{2}+\sqrt{18}-\sqrt{98}-\sqrt{288}\)

\(=\sqrt{2}\cdot\left(5+\sqrt{9}-\sqrt{49}-\sqrt{144}\right)\)

\(=\sqrt{2}\cdot\left(5+3-7-12\right)\)

\(=-11\sqrt{2}\)

e) Ta có: \(\left(\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{3}+\sqrt{5}}\)

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

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

\(=3-5=-2\)

g) Ta có: \(\left(\sqrt{3}-1\right)\cdot\sqrt{4+2\sqrt{3}}\)

\(=\left(\sqrt{3}-1\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\)

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

\(=\left(\sqrt{3}-1\right)\cdot\left|\sqrt{3}+1\right|\)

\(=\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1>0\))

\(=3-1=2\)

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