Упр.429 ГДЗ Макарычев Миндюк 8 класс (Алгебра)

1. Освободитесь от иррациональности в знаменателе дроби: а) $$\frac{x}{x+\sqrt{y}}$$; б) $$\frac{b}{a-\sqrt{b}}$$; в) $$\frac{4}{\sqrt{10}-\sqrt{2}}$$; г) $$\frac{12}{\sqrt{3}+\sqrt{6}}$$; д) $$\frac{9}{3-2\sqrt{2}}$$; е) $$\frac{14}{1+5\sqrt{2}}$$.
2. Сократите дробь:

1. а)

   $$\frac{x}{x+\sqrt{y}}=\frac{x(x-\sqrt{y})}{(x+\sqrt{y})(x-\sqrt{y})}=\frac{x(x-\sqrt{y})}{x^2-y}$$

2. б)

   $$\frac{b}{a-\sqrt{b}}=\frac{b(a+\sqrt{b})}{(a-\sqrt{b})(a+\sqrt{b})}=\frac{b(a+\sqrt{b})}{a^2-b}$$

3. в)

   $$\frac{4}{\sqrt{10}-\sqrt{2}}=\frac{4(\sqrt{10}+\sqrt{2})}{(\sqrt{10}-\sqrt{2})(\sqrt{10}+\sqrt{2})}=\frac{4(\sqrt{10}+\sqrt{2})}{10-2}=\frac{\sqrt{10}+\sqrt{2}}{2}$$

4. г)

   $$\frac{12}{\sqrt{3}+\sqrt{6}}=\frac{12(\sqrt{3}-\sqrt{6})}{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6})}=\frac{12(\sqrt{3}-\sqrt{6})}{3-6}=-4(\sqrt{3}-\sqrt{6})=4(\sqrt{6}-\sqrt{3})$$

5. д)

   $$\frac{9}{3-2\sqrt{2}}=\frac{9(3+2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}=\frac{9(3+2\sqrt{2})}{9-8}=9(3+2\sqrt{2})$$

6. е)

   $$\frac{14}{1+5\sqrt{2}}=\frac{14(1-5\sqrt{2})}{(1+5\sqrt{2})(1-5\sqrt{2})}=\frac{14(1-5\sqrt{2})}{1-50}=\frac{14(1-5\sqrt{2})}{-49}=\frac{2(5\sqrt{2}-1)}{7}$$

Ответ

а) $$\frac{x(x-\sqrt{y})}{x^2-y}$$; б) $$\frac{b(a+\sqrt{b})}{a^2-b}$$; в) $$\frac{\sqrt{10}+\sqrt{2}}{2}$$; г) $$4(\sqrt{6}-\sqrt{3})$$; д) $$9(3+2\sqrt{2})$$; е) $$\frac{2(5\sqrt{2}-1)}{7}$$.