Дополнительные упражнения к главе II — К параграфу 6 — 503 — стр. 114

\(\frac{1}{\sqrt{2}+\sqrt{3}+1}=\frac{\sqrt{2}+\sqrt{3}-1}{(\sqrt{2}+\sqrt{3}+1)(\sqrt{2}+\sqrt{3}-1)}=\frac{\sqrt{2}+\sqrt{3}-1}{(\sqrt{2}+\sqrt{3})^{2}-1}=\frac{\sqrt{2}+\sqrt{3}-1}{2+2 \sqrt{6}+3-1}=\frac{\sqrt{2}+\sqrt{3}-1}{4+2 \sqrt{6}}=\)

\(=\frac{(\sqrt{2}+\sqrt{3}-1)(2-\sqrt{6})}{2(2+\sqrt{6})(2-\sqrt{6})}=\frac{\sqrt{2}(\sqrt{2}+\sqrt{3}-1)(\sqrt{2}-\sqrt{3})}{-4}=-\frac{\sqrt{2}((\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})-(\sqrt{2}-\sqrt{3}))}{4}=\)

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

\(\frac{1}{\sqrt{5}-\sqrt{3}+2}=\frac{\sqrt{5}-\sqrt{3}-2}{(\sqrt{5}-\sqrt{3}+2)(\sqrt{5}-\sqrt{3}-2)}=\frac{\sqrt{5}-\sqrt{3}-2}{(\sqrt{5}-\sqrt{3})^{2}-4}=\frac{\sqrt{5}-\sqrt{3}-2}{5-2 \sqrt{15}+3-4}=\frac{\sqrt{5}-\sqrt{3}-2}{4-2 \sqrt{15}}=\)

\(=\frac{(\sqrt{5}-\sqrt{3}-2)(2+\sqrt{15})}{2(2-\sqrt{15})(2+\sqrt{15})}=\frac{2 \sqrt{5}-2 \sqrt{3}-4+\sqrt{75}-\sqrt{45}-2 \sqrt{15}}{2(4-15)}=-\frac{2 \sqrt{5}-2 \sqrt{3}-4+5 \sqrt{3}-3 \sqrt{5}-2 \sqrt{15}}{22}=\)

\(=-\frac{-\sqrt{5}+3 \sqrt{3}-4-2 \sqrt{15}}{22}=\frac{2 \sqrt{15}+\sqrt{5}-3 \sqrt{3}+4}{22}\).