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.

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

\(\sqrt{\left(2\sqrt{2-1}\right)^2}-\sqrt{17+12\sqrt{2}}\\ =\left|2\sqrt{2}-1\right|-\sqrt{9+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\\ =2\sqrt{2}-1-\sqrt{\left(3+2\sqrt{2}\right)^2}\\=2\sqrt{2}-1-\left(3+2\sqrt{2}\right)\\ =2\sqrt{2}-1-3-2\sqrt{2}\\ =-4\)

__

\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\\ =\left|2-\sqrt{5}\right|+\sqrt{9-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =2-\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}\\ =2-\sqrt{5}+3-\sqrt{5}\\ =5-2\sqrt{5}\)

__

\(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\\ =\left|4-3\sqrt{2}\right|-\sqrt{18+2\cdot3\cdot\sqrt{2}+1}\\ =4-3\sqrt{2}-\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =4-3\sqrt{2}-3\sqrt{2}-1\\ =3-6\sqrt{2}\)

a)

\((4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})^2=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})\)

\(=2(4^2-15)=2\)

b)

\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}=\sqrt{(8+2\sqrt{15})+2+2(\sqrt{6}+\sqrt{10})}\)

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

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

c)

\((\sqrt{5+2\sqrt{9\sqrt{5}-19}}-\sqrt{7-\sqrt{5}}):(2\sqrt{\sqrt{5}-2})\)

\(=(\sqrt{(5+2\sqrt{9\sqrt{5}-19})(\sqrt{5}+2)}-\sqrt{(7-\sqrt{5})(\sqrt{5}+2)}):(2\sqrt{(\sqrt{5}-2)(\sqrt{5}+2)})\)

\(=[\sqrt{10+5\sqrt{5}+2\sqrt{(9\sqrt{5}-19)(9+4\sqrt{5})}}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{10+5\sqrt{5}+2\sqrt{9+5\sqrt{5}}}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{(9+5\sqrt{5})+2\sqrt{9+5\sqrt{5}}+1}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{(\sqrt{9+5\sqrt{5}}+1)^2}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{9+5\sqrt{5}}+1-\sqrt{9+5\sqrt{5}}]:2=\frac{1}{2}\)

d)

\((\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}})^2=18+2\sqrt{(9+\sqrt{5})(9-\sqrt{5})}=18+4\sqrt{19}\)

\(\Rightarrow \sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}=\sqrt{18+4\sqrt{19}}\)

Do đó:
\(\frac{\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{3-2\sqrt{2}}=\frac{\sqrt{18+4\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{2+1-2\sqrt{2.1}}\)

\(=\frac{\sqrt{2}.\sqrt{9+2\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-(\sqrt{2}-1)=1\)

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

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

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

\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)

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

a: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)

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

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

b: \(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

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

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

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

c: \(\dfrac{3}{2\sqrt{3}+3}+\dfrac{3}{2\sqrt{3}-3}\)

\(=\dfrac{3\left(2\sqrt{3}-3\right)+3\left(2\sqrt{3}+3\right)}{12-9}\)

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

d: \(\sqrt{\left(\sqrt{3}+4\right)\cdot\sqrt{19-8\sqrt{3}}+3}\)

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

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

\(=\sqrt{16-3+3}=\sqrt{16}=4\)

e: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)

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

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

1: \(=\sqrt{36}=6\)

2: \(=\sqrt{\left(15-9\right)\left(15+9\right)}=\sqrt{24\cdot6}=12\)

3: \(=3\sqrt{5}-1-3\sqrt{5}-1=-2\)

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

5: \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)