Упр.112 ГДЗ Алимов 10-11 класс (Алгебра)

Рассмотрим вариант решения задания из учебника Алимов, Колягин, Ткачёва 10 класс, Просвещение: 112. Освободиться от иррациональности в знаменателе дроби:

Решение

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

2. $$\frac{\sqrt{5}}{5+\sqrt{10}}=\frac{\sqrt{5}(5-\sqrt{10})}{(5+\sqrt{10})(5-\sqrt{10})}=\frac{\sqrt{5}(5-\sqrt{10})}{25-10}=\frac{\sqrt{5}(5-\sqrt{10})}{15}$$

   $$\frac{\sqrt{5}(5-\sqrt{10})}{15}=\frac{5\sqrt{5}-\sqrt{50}}{15}=\frac{5\sqrt{5}-5\sqrt{2}}{15}=\frac{\sqrt{5}-\sqrt{2}}{3}$$

3. $$\frac{3}{\sqrt[3]{4}}=\frac{3\sqrt[3]{2}}{\sqrt[3]{4}\cdot \sqrt[3]{2}}=\frac{3\sqrt[3]{2}}{\sqrt[3]{8}}=\frac{3\sqrt[3]{2}}{2}$$

4. $$\frac{2}{\sqrt[4]{27}}=\frac{2\sqrt[4]{3}}{\sqrt[4]{27}\cdot \sqrt[4]{3}}=\frac{2\sqrt[4]{3}}{\sqrt[4]{81}}=\frac{2\sqrt[4]{3}}{3}$$

5. $$\frac{3}{\sqrt[4]{5}-\sqrt[4]{2}}=\frac{3(\sqrt[4]{5}+\sqrt[4]{2})}{(\sqrt[4]{5}-\sqrt[4]{2})(\sqrt[4]{5}+\sqrt[4]{2})}=\frac{3(\sqrt[4]{5}+\sqrt[4]{2})}{\sqrt{5}-\sqrt{2}}$$

   $$\frac{3(\sqrt[4]{5}+\sqrt[4]{2})}{\sqrt{5}-\sqrt{2}}=\frac{3(\sqrt[4]{5}+\sqrt[4]{2})(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}=(\sqrt[4]{5}+\sqrt[4]{2})(\sqrt{5}+\sqrt{2})$$

6. $$\frac{11}{\sqrt[3]{3}+\sqrt[3]{2}}=\frac{11\left((\sqrt[3]{3})^2-\sqrt[3]{3}\cdot \sqrt[3]{2}+(\sqrt[3]{2})^2\right)}{(\sqrt[3]{3}+\sqrt[3]{2})\left((\sqrt[3]{3})^2-\sqrt[3]{3}\cdot \sqrt[3]{2}+(\sqrt[3]{2})^2\right)}$$

   $$=\frac{11(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4})}{3+2}=\frac{11(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4})}{5}$$

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

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

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

8. $$\frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}=\frac{\sqrt[3]{3}-\sqrt[3]{2}}{(\sqrt[3]{3}-\sqrt[3]{2})\left((\sqrt[3]{2})^2+\sqrt[3]{2}\cdot \sqrt[3]{3}+(\sqrt[3]{3})^2\right)}$$

   $$=\frac{\sqrt[3]{3}-\sqrt[3]{2}}{3-2}=\sqrt[3]{3}-\sqrt[3]{2}$$

Ответ

1. $$-2(\sqrt{2}+\sqrt{3})$$
2. $$\frac{\sqrt{5}-\sqrt{2}}{3}$$
3. $$\frac{3\sqrt[3]{2}}{2}$$
4. $$\frac{2\sqrt[4]{3}}{3}$$
5. $$\left(\sqrt[4]{5}+\sqrt[4]{2}\right)\left(\sqrt{5}+\sqrt{2}\right)$$
6. $$\frac{11(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4})}{5}$$
7. $$\frac{\sqrt{2}+2-\sqrt{6}}{4}$$
8. $$\sqrt[3]{3}-\sqrt[3]{2}$$