Задание 6 Избавьтесь от иррациональности: 5 1 √3 a) √12; √2+1 √2 √5 б) √11-√3; 3√2-1 √5 2 1

Задание 6

Избавьтесь от иррациональности:

a) $$\frac{5}{\sqrt{12}}; \frac{1}{\sqrt{2}+1}; \frac{\sqrt{3}}{\sqrt{2} \cdot \sqrt{5}}$$

1. $$\frac{5}{\sqrt{12}} = \frac{5}{\sqrt{4 \cdot 3}} = \frac{5}{2\sqrt{3}} = \frac{5\sqrt{3}}{2 \cdot 3} = \frac{5\sqrt{3}}{6}$$
2. $$\frac{1}{\sqrt{2}+1} = \frac{1(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{\sqrt{2}-1}{2 - 1} = \sqrt{2} - 1$$
3. $$\frac{\sqrt{3}}{\sqrt{2} \cdot \sqrt{5}} = \frac{\sqrt{3}}{\sqrt{10}} = \frac{\sqrt{3} \cdot \sqrt{10}}{10} = \frac{\sqrt{30}}{10}$$

Ответ: $$\frac{5\sqrt{3}}{6}; \sqrt{2} - 1; \frac{\sqrt{30}}{10}$$

б) $$\frac{2}{\sqrt{11}-\sqrt{3}}; \frac{1}{3\sqrt{2}-1}; \frac{\sqrt{5}}{1}$$

1. $$\frac{2}{\sqrt{11}-\sqrt{3}} = \frac{2(\sqrt{11}+\sqrt{3})}{(\sqrt{11}-\sqrt{3})(\sqrt{11}+\sqrt{3})} = \frac{2(\sqrt{11}+\sqrt{3})}{11-3} = \frac{2(\sqrt{11}+\sqrt{3})}{8} = \frac{\sqrt{11}+\sqrt{3}}{4}$$
2. $$\frac{1}{3\sqrt{2}-1} = \frac{1(3\sqrt{2}+1)}{(3\sqrt{2}-1)(3\sqrt{2}+1)} = \frac{3\sqrt{2}+1}{18-1} = \frac{3\sqrt{2}+1}{17}$$
3. $$\frac{\sqrt{5}}{1} = \sqrt{5}$$

Ответ: $$\frac{\sqrt{11}+\sqrt{3}}{4}; \frac{3\sqrt{2}+1}{17}; \sqrt{5}$$