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

Освободиться от иррациональности в знаменателе дроби:

1)

$$\frac{2}{\sqrt2-\sqrt3}=\frac{2(\sqrt2+\sqrt3)}{(\sqrt2-\sqrt3)(\sqrt2+\sqrt3)}=\frac{2(\sqrt2+\sqrt3)}{2-3}=-2(\sqrt2+\sqrt3).$$

Ответ: $$-2(\sqrt2+\sqrt3)$$

2)

$$\frac{\sqrt5}{5+\sqrt{10}}=\frac{\sqrt5(5-\sqrt{10})}{(5+\sqrt{10})(5-\sqrt{10})}=\frac{\sqrt5(5-\sqrt{10})}{25-10}=\frac{\sqrt5(5-\sqrt{10})}{15}$$

$$=\frac{5\sqrt5-\sqrt{50}}{15}=\frac{5\sqrt5-5\sqrt2}{15}=\frac{\sqrt5-\sqrt2}{3}.$$

Ответ: $$\frac{\sqrt5-\sqrt2}{3}$$

3)

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

Ответ: $$\frac{3\sqrt[3]{2}}{2}$$

4)

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

Ответ: $$\frac{2\sqrt[4]{3}}{3}$$

5)

$$\frac{3}{\sqrt[4]{5}-\sqrt[4]{2}}=\frac{3(\sqrt[4]{5}+\sqrt[4]{2})}{(\sqrt[4]{5}-\sqrt[4]{2})(\sqrt[4]{5}+\sqrt[4]{2})}=\frac{3(\sqrt[4]{5}+\sqrt[4]{2})}{\sqrt5-\sqrt2}$$

$$=\frac{3(\sqrt[4]{5}+\sqrt[4]{2})(\sqrt5+\sqrt2)}{(\sqrt5-\sqrt2)(\sqrt5+\sqrt2)}=(\sqrt[4]{5}+\sqrt[4]{2})(\sqrt5+\sqrt2).$$

Ответ: $$(\sqrt[4]{5}+\sqrt[4]{2})(\sqrt5+\sqrt2)$$

6)

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

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

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

7)

$$\frac{1}{1+\sqrt2+\sqrt3}=\frac{1+\sqrt2-\sqrt3}{(1+\sqrt2+\sqrt3)(1+\sqrt2-\sqrt3)}$$

$$=\frac{1+\sqrt2-\sqrt3}{(1+\sqrt2)^2-(\sqrt3)^2}=\frac{1+\sqrt2-\sqrt3}{1+2\sqrt2+2-3}=\frac{1+\sqrt2-\sqrt3}{2\sqrt2}$$

$$=\frac{\sqrt2+2-\sqrt6}{4}.$$

Ответ: $$\frac{\sqrt2+2-\sqrt6}{4}$$

8)

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

$$=\frac{\sqrt[3]{3}-\sqrt[3]{2}}{(\sqrt[3]{3})^3-(\sqrt[3]{2})^3}=\frac{\sqrt[3]{3}-\sqrt[3]{2}}{3-2}=\sqrt[3]{3}-\sqrt[3]{2}.$$

Ответ: $$\sqrt[3]{3}-\sqrt[3]{2}$$