(m^1/2n^-2-n^-3/2) делить m^3/4-m^1/4n^1/2 при - вопрос №5805871122323414128101 от rope22 22.05.2023 14:32

Question

Алгебра: (m^1/2n^-2-n^-3/2) делить m^3/4-m^1/4n^1/2 при m=81 n=0,1

rope22 · Answer

(m^1/2 n^(-2) - n^(-3/2)) / (m^3/4 - m^1/4 n^1/2) = (m^1/2 : n^2 - 1:n^3/2) / /m^1/4(m^1/2 - n^1/2) = (m^1/2 - n^1/2) / n^2 * 1 / m^1/4(m^1/2 - n^1/2) = = 1 : (n^2 * m^1/4) при m = 81, n = 0.1 Получим
1 : (0.1^2 * 81^1/4) = 1/ 0.03 = 100/3 = 33 1/3