a: Ta có: \(\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}\)

\(=5-\sqrt{19}-\sqrt{19}+4\)

\(=9-2\sqrt{19}\)

b: Ta có: \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-3\right)^2}\)

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

=0

c.

Căn bậc 2 không xác định do $2-\sqrt{5}< 0$

d.

\(=\sqrt{(3+\sqrt{3})^2}(3+\sqrt{3})=|3+\sqrt{3}|(3+\sqrt{3})=(3+\sqrt{3})^2=12+6\sqrt{3}\)

e.

\(=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1\)

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

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

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

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

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

\(=3-1\)

\(=2\)

b, \(\left(\sqrt{4+\sqrt{15}}-\sqrt{16-3\sqrt{15}}\right)\left(\sqrt{3}+\sqrt{5}\right)\)

\(=\frac{\sqrt{8+2\sqrt{15}}-\sqrt{32-6\sqrt{15}}}{\sqrt{2}}.\left(\sqrt{3}+\sqrt{5}\right)\)

\(=\frac{\sqrt{3+2\sqrt{3}.\sqrt{5}+5}-\sqrt{27-2.3\sqrt{3}.\sqrt{5}+5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)

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

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

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

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

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

\(=2\sqrt{2}\)

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

\(=\left(\frac{5+2\sqrt{6}+2\left(5-2\sqrt{6}\right)}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\cdot\left(15+2\sqrt{6}\right)\)

\(=\frac{5+2\sqrt{6}+10-4\sqrt{6}}{25-24}\cdot\left(15+2\sqrt{6}\right)\)

\(=\left(15-2\sqrt{6}\right)\cdot\left(15+2\sqrt{6}\right)\)

\(=15^2-\left(2\sqrt{6}\right)^2\)

\(=225-24=201\)

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

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

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

2: \(=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)

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

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

3: \(=\left(\sqrt{7}-\sqrt{2}+\sqrt{7}+\sqrt{2}\right)^2=\left(2\sqrt{7}\right)^2=28\)