x²/3>8x-9/5
x²/3-(8x+9)/5>0 /*15
5x²-24x+27>0
5x²-24x+27=0
D = 576-540 = 36 = 6²
x₁ = (24+6):(5*2) = 30:10 = 3
x₂ = (24-6):(5*2) = 18:10 = 1,8
5x²-24x+27= (x-1,8)*(x-3)
(x-1,8)*(x-3)>0
+ - +
_________________> x
× ×
1,8 3
x∈(-∞;1,8)U(3;+∞)
Ответ: x∈(-∞;1,8)U(3;+∞).