Упр.23.12 ГДЗ Мордкович Семенов 8 класс (Алгебра)

1. Сократите дробь: а) $$\frac{1-\sqrt{3}}{\sqrt{2}-\sqrt{6}}$$; б) $$\frac{\sqrt{5}+\sqrt{10}}{1+\sqrt{2}}$$; в) $$\frac{4a+4\sqrt{3}}{3-a^2}$$; г) $$\frac{\sqrt{6}-\sqrt{10}}{\sqrt{15}-5}$$; д) $$\frac{\sqrt{21}-\sqrt{6}}{7-\sqrt{14}}$$; е) $$\frac{x-25}{3\sqrt{x}+15}$$.

1. а)
   $$\frac{1-\sqrt{3}}{\sqrt{2}-\sqrt{6}}=\frac{1-\sqrt{3}}{\sqrt{2}-\sqrt{2}\sqrt{3}}=\frac{1-\sqrt{3}}{\sqrt{2}(1-\sqrt{3})}=\frac{1}{\sqrt{2}}.$$

2. б)
   $$\frac{\sqrt{5}+\sqrt{10}}{1+\sqrt{2}}=\frac{\sqrt{5}+\sqrt{5}\sqrt{2}}{1+\sqrt{2}}=\frac{\sqrt{5}(1+\sqrt{2})}{1+\sqrt{2}}=\sqrt{5}.$$

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

4. г)
   $$\frac{\sqrt{6}-\sqrt{10}}{\sqrt{15}-5}=\frac{\sqrt{2}\sqrt{3}-\sqrt{2}\sqrt{5}}{\sqrt{5}\sqrt{3}-\sqrt{5}\sqrt{5}}=\frac{\sqrt{2}(\sqrt{3}-\sqrt{5})}{\sqrt{5}(\sqrt{3}-\sqrt{5})}=\frac{\sqrt{2}}{\sqrt{5}}.$$

5. д)
   $$\frac{\sqrt{21}-\sqrt{6}}{\sqrt{7}-\sqrt{14}}=\frac{\sqrt{3}\sqrt{7}-\sqrt{3}\sqrt{2}}{\sqrt{7}\sqrt{7}-\sqrt{7}\sqrt{2}}=\frac{\sqrt{3}(\sqrt{7}-\sqrt{2})}{\sqrt{7}(\sqrt{7}-\sqrt{2})}=\frac{\sqrt{3}}{\sqrt{7}}.$$

6. е)
   $$\frac{x-25}{3\sqrt{x}+15}=\frac{(\sqrt{x}-5)(\sqrt{x}+5)}{3(\sqrt{x}+5)}=\frac{\sqrt{x}-5}{3}.$$

Ответ

а) $$\frac{1}{\sqrt{2}}$$; б) $$\sqrt{5}$$; в) $$\frac{4}{\sqrt{3}-a}$$; г) $$\frac{\sqrt{2}}{\sqrt{5}}$$; д) $$\frac{\sqrt{3}}{\sqrt{7}}$$; е) $$\frac{\sqrt{x}-5}{3}$$.