Răspuns:
[tex](x-3)(x-6) < (x-1)(x-2)\\ \\(x-3)(x-6)=x^2-3x-6x+18=x^2-9x+18\\ \\(x-1)(x-2)=x^2-x-2x+2=x^2-3x+2\\\\x^2-9x+18 < x^2-3x+2x^2-9x+18-x^2+3x-2 < 0\\\\-6x+16 < 0 /*(-1) \\\\6x-16 > 0= > 6x > 16= > x > 16/6= > x > 8/3[/tex]
x ∈ ( 8/3 , ∞)
