Решите уравнение 7*49^x+3*28^x = 4*16^x

7*49^x+3*28^x = 4*16^x7*7^(2x)+3*7^x*4^x - 4*4^(2x) = 0 | : 4^(2x)≠07*(7/4)^(2x)+3*(7/4)^x - 4 = 0замена: (7/4)^x = t > 07*t² + 3*t - 4 = 0D=9+4*7*4 = 11²t₁ = (-3-11)/14 = -1 посторонний кореньt₂ = (-3+11)/14 = 4/7x = -1