Освобождение от иррациональности в знаменателе дроби

• a) $$\frac{4}{\sqrt{3}+1} = \frac{4(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{4(\sqrt{3}-1)}{3-1} = \frac{4(\sqrt{3}-1)}{2} = 2(\sqrt{3}-1) = 2\sqrt{3} - 2$$


• б) $$\frac{1}{1-\sqrt{2}} = \frac{1(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})} = \frac{1+\sqrt{2}}{1-2} = \frac{1+\sqrt{2}}{-1} = -1-\sqrt{2}$$


• в) $$\frac{1}{\sqrt{x}-\sqrt{y}} = \frac{1(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})} = \frac{\sqrt{x}+\sqrt{y}}{x-y}$$


• г) $$\frac{33}{7-3\sqrt{3}} = \frac{33(7+3\sqrt{3})}{(7-3\sqrt{3})(7+3\sqrt{3})} = \frac{33(7+3\sqrt{3})}{49-9 \cdot 3} = \frac{33(7+3\sqrt{3})}{49-27} = \frac{33(7+3\sqrt{3})}{22} = \frac{3(7+3\sqrt{3})}{2} = \frac{21+9\sqrt{3}}{2}$$


• e) $$\frac{15}{2\sqrt{5}+5} = \frac{15(2\sqrt{5}-5)}{(2\sqrt{5}+5)(2\sqrt{5}-5)} = \frac{15(2\sqrt{5}-5)}{4 \cdot 5 - 25} = \frac{15(2\sqrt{5}-5)}{20-25} = \frac{15(2\sqrt{5}-5)}{-5} = -3(2\sqrt{5}-5) = -6\sqrt{5}+15 = 15-6\sqrt{5}$$