Упр.549 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

1) 1/(3-v2)+1/(3+v2); 2) 1/(5-v3)-1/(5+v3);
3) (3-v2)/(3+v2)-(3+v2)/(3-v2); 4) 3/(v3-v2)-3/(v3+v2).

1. $$\frac{1}{3-\sqrt{2}}+\frac{1}{3+\sqrt{2}}= \frac{(3+\sqrt{2})+(3-\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}= \frac{6}{9-2}=\frac{6}{7}.$$

2. $$\frac{1}{5-\sqrt{3}}-\frac{1}{5+\sqrt{3}}= \frac{(5+\sqrt{3})-(5-\sqrt{3})}{(5-\sqrt{3})(5+\sqrt{3})}= \frac{2\sqrt{3}}{25-3}=\frac{2\sqrt{3}}{22}=\frac{\sqrt{3}}{11}.$$

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

   $$\frac{(9-6\sqrt{2}+2)-(9+6\sqrt{2}+2)}{9-2}= \frac{-12\sqrt{2}}{7}.$$

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

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

Ответ

$$1)\ \frac{6}{7};\quad 2)\ \frac{\sqrt{3}}{11};\quad 3)\ -\frac{12\sqrt{2}}{7};\quad 4)\ 6\sqrt{2}.$$