\frac{1}{\sqrt2+\sqrt3-1}=
=\frac{1}{[\sqrt2+(\sqrt3-1)]}*\frac{[\sqrt2-(\sqrt3-1]}{[\sqrt2-(\sqrt3-1)]}=
=\frac{\sqrt2-\sqrt3+1}{\sqrt2^2-(\sqrt3-1)^2}=
=\frac{\sqrt2-\sqrt3+1}{2-(3-2\sqrt3+1)}=
=\frac{\sqrt2-\sqrt3+1}{2-3+2\sqrt3-1}=
=\frac{\sqrt2-\sqrt3+1}{{2\sqrt3-2}}=
=\frac{\sqrt2-\sqrt3+1}{2(\sqrt3-1)}*\frac{\sqrt3+1}{\sqrt3+1}=
=\frac{\sqrt6+\sqrt2-3-\sqrt3+\sqrt3+1}{2(3-1)}=
=\frac{\sqrt6+\sqrt2-2}{4}=\frac{1}{4}(\sqrt6+\sqrt2-2)