\frac{3}{1+\sqrt2}+\frac{3}{\sqrt2+\sqrt3}+\frac{3}{\sqrt3+2}=
\frac{3(1-\sqrt2)}{(1+\sqrt2)(1-\sqrt2)}+\frac{3(\sqrt2-\sqrt3)}{(\sqrt2+\sqrt3)(\sqrt2-\sqrt3)}+\frac{3(\sqrt3-2)}{(\sqrt3+2)(\sqrt3-2)}=
\frac{3(1-\sqrt2)}{1-2}+\frac{3(\sqrt2-\sqrt3)}{2-3}+\frac{3(\sqrt3-2)}{3-4}=
\frac{3(1-\sqrt2)}{-1}+\frac{3(\sqrt2-\sqrt3)}{-1}+\frac{3(\sqrt3-2)}{-1}=
-3(1-\sqrt2)-3(\sqrt2-\sqrt3)-3(\sqrt3-2)=
-3+3\sqrt2-3\sqrt2+3\sqrt3-3\sqrt3+6=3
odpowiedź C