Решите неравенство 1 + sqrt(_9(3x^2 + 8x + 6)) > _3(3x^2 + 8x + 6).

Положим y = sqrt(_9(3x^2 + 8x + 6)) . Тогда 1 + sqrt(_9(3x^2 + 8x + 6)) > _3(3x^2 + 8x + 6) 1 + y > 2y^2 (2y+1)(y-1) < 0 -(1)/(2) < y < 1 0 _9(3x^2 + 8x + 6) < 1 1 3x^2 + 8x + 6 < 9 cases 3x^2 + 8x + 5 0 3x^2 + 8x - 3 < 0 cases cases [arrayl x -(5)/(3) x -1 array. -3 < x < (1)/(3) cases x in (-3, -(5)/(3)] U [-1, (1)/(3))

\( x \in \left(-3, -\frac{5}{3}\right] \cup \left[-1, \frac{1}{3}\right) \)