6 и 7 умоляю !!! Быстрей , помогите !!!!
6) (1 - 2y)^2 = y(y + 3) - 1 1 - 4y + 4y^2 = y^2 + 3y - 1 3y^2 - 7y + 2 = 0 (y - 2)(3y - 1) = 0 y1 = 2, y2 = 1/3 7) x(x + 1) = x + x + 1 + 11 x^2 + x = 2x + 12 x^2 - x - 12 = 0 (x + 3)(x - 4) = 0 x1 = -3 < 0 - не натуральное, x2 = 4