a) \(\sqrt{11-6\sqrt{2}}-\sqrt{27+10\sqrt{2}}\)

\(=\sqrt{9-6\sqrt{2}+2}-\sqrt{25+10\sqrt{2}+2}\)

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

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

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

b) \(\sqrt{13-4\sqrt{3}}-\sqrt{16-8\sqrt{3}}\)

\(=\sqrt{12-4\sqrt{3}+1}-\sqrt{12-8\sqrt{3}+4}\)

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

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

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

\(=1\)

a: Ta có: \(\sqrt{75}-2\sqrt{27}+\sqrt{48}\)

\(=5\sqrt{3}-2\cdot3\sqrt{3}+4\sqrt{3}\)

\(=3\sqrt{3}\)

c: Ta có: \(\sqrt{8+2\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)

\(=\sqrt{7}+1-\sqrt{7}+2\)

=3

\(a.\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=2\sqrt{5}+\dfrac{8}{1-\sqrt{5}}=\dfrac{2\sqrt{5}-2}{1-\sqrt{5}}=\dfrac{-2\left(1-\sqrt{5}\right)}{1-\sqrt{5}}=-2\) \(b.\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6.5}+\sqrt{27.6}}=\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{17}\right)}=-\dfrac{\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-2-1}{\sqrt{6}}=-\dfrac{\sqrt{3}}{\sqrt{2}}\)

a) \(\sqrt{3+2\sqrt{2}}-\sqrt{17-12\sqrt{2}}\)

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

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

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

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

b) \(\sqrt{5-2\sqrt{6}}-\sqrt{14-4\sqrt{6}}-\sqrt{48}\)

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

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

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

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

c) \(\sqrt{11+3\sqrt{8}}-\sqrt{17-12\sqrt{2}}-4\sqrt{8}\)

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

= \(\left|3+\sqrt{2}\right|-\left|3-2\sqrt{2}\right|-8\sqrt{2}\)

= \(3+\sqrt{2}-3+2\sqrt{2}-8\sqrt{2}\)

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

\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)

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

1) \(=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\)

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

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

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

a) Ta có: \(-3\sqrt{16}\cdot\sqrt{90}\)

\(=-3\cdot4\cdot3\sqrt{10}\)

\(=-36\sqrt{10}\)

b) Ta có: \(3\sqrt{\dfrac{4}{3}}-3\sqrt{48}+5\sqrt{75}\)

\(=3\cdot\dfrac{2}{\sqrt{3}}-3\cdot4\sqrt{3}+5\cdot5\sqrt{3}\)

\(=2\sqrt{3}-12\sqrt{3}+25\sqrt{3}\)

\(=15\sqrt{3}\)

c) Ta có: \(4\sqrt[3]{27}-\sqrt[3]{64}-2\sqrt[3]{8}\)

\(=4\cdot3-4-2\cdot2\)

\(=12-4-4=4\)