1)2√2+√50-√98 2)√16+2√40с-3√90с 3)√3(4√3-2√6) - вопрос №3502583122509821624039033432645725725237 от sanekmmg1980 24.11.2020 00:29

Question

Алгебра: 1)2√2+√50-√98 2)√16+2√40с-3√90с 3)√3(4√3-2√6) 4)(5√7-2)(5√7+2) 5)(2√3-7)^2

sanekmmg1980 · Answer

1)2√2+√50 - √98 = √2 + 5√2 - 7√2 = - √22)√16+ 2√40с - 3√90с = 4 + 2√(4*10)с - 3√(9*10)с = 4 + 4с√10 - 9с√10 = = 4 - 5√103)√3(4√3-2√6) = 4*3 - 2√(9*2) = 12 - 6√24)(5√7-2)(5√7+2) = (5√7)² - 2² = 25*7 - 4 = 1715)(2√3-7)² = (2√3)² - 28√3 + 49 = 12 - 28√3 + 49 = 37 - 28√3...