Question

0,7(х+6)=0,6(х-4)+12,6

Answer 1

Log0,5(x^6 - 6x^4 + 12x^2 - 8) = -3
 
 x^6 - 6x^4 + 12x^2 - 8 = (1/2)^-3
 x^6 - 6x^4 + 12x^2 - 8 = 8
 x^6 - 6x^4 + 12x^2 - 16 = 0
 Замена переменных
 t = x^2
 t^3 - 6^2t + 12t - 16 = 0
 (t^3 - 3*2*t^2 + 3*2^2*t + (-2)^3)) - 8 = 0
 (t - 2)^3 - 8 = 0
 (t - 2)^3 - 2^3 = 0
 (t - 4)(t^2 - 2t + 4) = 0
 t - 4 = 0   t^2 - 2t + 4 = 0
 t = 4      D = (-2)^2 - 4*4 = 4 - 16 = -12 
 Находим значения х
 x^2 = 4
 x1 = 2
 x2 = -2

Answer 2

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