Решите уравнение; (4+x)^7=-128; (10x)^2=900; 3^3k=729; - вопрос №77649044712810290033729752343 от likagluxova 18.09.2022 09:27

Question

Алгебра: Решите уравнение; (4+x)^7=-128; (10x)^2=900; 3^3k=729; 7^5n^-2=343

likagluxova · Answer

(4+x)^7 = -128(4+x)^7 = (-2)^74 + х = -2х = -2/4х = -1/2х = -0,5 (10x)^2=90010х = ± √90010х = ± 30▪х1 = 30÷10х1 = 3▪х2 = -30÷10х2 = -33^(3k) = 7293^(3k) = 3^63k = 6k = 6÷3k = 2(7^5)(n^-2)=343. ОДЗ: n≠07^5/n^2 = 343343n^2 = 7^5(7^3)(n^2) = 7^5n^2 = 7^5/7^3n^2 = 7^2n = ± 7...