Решите ! log2 (4x-3)+log1/8 125 = log1/2 x + - вопрос №345206624318125124004 от natalyakuznecova81 21.12.2021 05:36

Question

Алгебра: Решите ! log2 (4x-3)+log1/8 125 = log1/2 x + log4 0,04

natalyakuznecova81 · Answer

Доп. условие x (3/4)log2 (4x-3) - (1/3)log2 (125) = -log2 (x) + (1/2)log2 (1/25)log2 (4x-3) - log2 (125)^(1/3) = -log2 (x) + log2 (1/5)log2 ((4x-3)/5) = - log2 5x(4x-3)/5=1/(5x)4x^2-3x-1=0x=1 и x= - (1/4) -не удовлетворяет по доп условиюответ: 1...