[tex]\bold{17.} \ (\sqrt{10-\sqrt{19}}+\sqrt{10+\sqrt19}})^2= \\ =(\sqrt{10-\sqrt{19}})^2+2\sqrt{10-\sqrt{19}}\cdot \sqrt{10+\sqrt{19}}+(\sqrt{10+\sqrt{19}})^2= \\ =10-\sqrt{19}+2\sqrt{(10-\sqrt{19})(10+\sqrt{19})}+10+\sqrt{19}=\\=20+2\sqrt{10^2-(\sqrt{19})^2}=20+2\sqrt{100-19}=20+2\sqrt{81}=20+2\cdot 9= \boxed{38};[/tex]
[tex]\bold{18.} \\ a) \ \sqrt3\cdot \sqrt{3+\sqrt6}\cdot \sqrt{3-\sqrt6}=3; \\ \sqrt{3(3+\sqrt6)(3-\sqrt6)}=\sqrt{3(3^2-(\sqrt6)^2)}=\sqrt{3(9-6)}=\sqrt{3 \cdot 3}=\boxed{3};[/tex]
[tex]c)\ \sqrt{2+\sqrt3}=\frac{\sqrt3-1}{\sqrt{2}} \\ \frac{\sqrt3-1}{\sqrt2}=\frac{\sqrt2\cdot(\sqrt3-1)}{(\sqrt2)^2}=\frac{\sqrt6-\sqrt2}{2} \\ \Rightarrow \boxed{\sqrt{2+\sqrt3}\neq\frac{\sqrt3-1}{\sqrt{2}} }[/tex]
