Упр.722 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

1. Освободиться от иррациональности в знаменателе дроби: 1) $$\frac{5}{\sqrt{3}-\sqrt{2}}$$; 2) $$\frac{3}{\sqrt{6}+\sqrt{5}}$$; 3) $$\frac{12}{\sqrt{10}-\sqrt{7}}$$; 4) $$\frac{8}{\sqrt{11}+\sqrt{3}}$$.

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

2. $$\frac{3}{\sqrt{6}+\sqrt{5}}=\frac{3(\sqrt{6}-\sqrt{5})}{(\sqrt{6}+\sqrt{5})(\sqrt{6}-\sqrt{5})}=\frac{3(\sqrt{6}-\sqrt{5})}{6-5}=3(\sqrt{6}-\sqrt{5}).$$

3. $$\frac{12}{\sqrt{10}-\sqrt{7}}=\frac{12(\sqrt{10}+\sqrt{7})}{(\sqrt{10}-\sqrt{7})(\sqrt{10}+\sqrt{7})}=\frac{12(\sqrt{10}+\sqrt{7})}{10-7}=4(\sqrt{10}+\sqrt{7}).$$

4. $$\frac{8}{\sqrt{11}+\sqrt{3}}=\frac{8(\sqrt{11}-\sqrt{3})}{(\sqrt{11}+\sqrt{3})(\sqrt{11}-\sqrt{3})}=\frac{8(\sqrt{11}-\sqrt{3})}{11-3}=\sqrt{11}-\sqrt{3}.$$

Ответ

1) $$5(\sqrt{3}+\sqrt{2})$$; 2) $$3(\sqrt{6}-\sqrt{5})$$; 3) $$4(\sqrt{10}+\sqrt{7})$$; 4) $$\sqrt{11}-\sqrt{3}$$.