Адание 5. Найдите зн 1 (√20-√5)√5; 2 (√18-√2)√2; 3 (√48-√3)√3; 4 (√50+√2)√2; 5 (√45+√5).√5; 6 (√27+√3) √3;

1. $$(\sqrt{20}-\sqrt{5})\cdot \sqrt{5} = \sqrt{20}\cdot \sqrt{5} - \sqrt{5}\cdot \sqrt{5} = \sqrt{100} - 5 = 10-5 = 5$$

2. $$(\sqrt{18}-\sqrt{2})\cdot \sqrt{2} = \sqrt{18}\cdot \sqrt{2} - \sqrt{2}\cdot \sqrt{2} = \sqrt{36} - 2 = 6-2 = 4$$

3. $$(\sqrt{48}-\sqrt{3})\cdot \sqrt{3} = \sqrt{48}\cdot \sqrt{3} - \sqrt{3}\cdot \sqrt{3} = \sqrt{144} - 3 = 12-3 = 9$$

4. $$(\sqrt{50}+\sqrt{2})\cdot \sqrt{2} = \sqrt{50}\cdot \sqrt{2} + \sqrt{2}\cdot \sqrt{2} = \sqrt{100} + 2 = 10+2 = 12$$

5. $$(\sqrt{45}+\sqrt{5})\cdot \sqrt{5} = \sqrt{45}\cdot \sqrt{5} + \sqrt{5}\cdot \sqrt{5} = \sqrt{225} + 5 = 15+5 = 20$$

6. $$(\sqrt{27}+\sqrt{3})\cdot \sqrt{3} = \sqrt{27}\cdot \sqrt{3} + \sqrt{3}\cdot \sqrt{3} = \sqrt{81} + 3 = 9+3 = 12$$

Ответ: 1) 5; 2) 4; 3) 9; 4) 12; 5) 20; 6) 12