Найдите произведение уравнения log² 0.2(x/25)+log² 0.2(x/5)=1

1) log₀₎₂²(x/25) = log₀₎₂(x/25)*log₀₎₂(x/25) =  = ( log₀₎₂x - log₀₎₂25)*( log₀₎₂x - log₀₎₂25)= (log₀₎₂x  +2)* (log₀₎₂x  +2)= = (log₀₎₂x  +2)²= log₀₎₂²x +4log₀₎₂x +4 2)log₀₎₂²(x/5) = log₀₎₂(x/5)*log₀₎₂(x/5) = (log₀₎₂x - log₀₎₂5)*(log₀₎₂x - log₀₎₂5)= =(log₀₎₂x +1)*(log₀₎₂x +1)= (log₀₎₂x +1)² = log₀₎₂²x + 2log₀₎₂x +1 3) Само уравнение:  log₀₎₂²x +4log₀₎₂x +4 +log₀₎₂²x + 2log₀₎₂x +1 = 1   (ОДЗ: x > 0) log₀₎₂x = t t² +4t +4 +t² +2t = 0 2t² +6t +4 = 0 t² +3t +2 = 0 По т. Виета  а) t = -2, ⇒ log₀₎₂x = -2, x = 0,2⁻² = 25 б) t = -1, ⇒ log₀₎₂x = -1, ⇒ x = 0,2⁻¹ = 5 Ответ: 125