Cześć!
[tex](x-3)^2>x-3\\x^2-6x+9>x-3\\x^2-7x+12>0\\[/tex]
Δ = [tex]49 - (4*1*12)=49-48=1[/tex]
[tex]x_{1} = \frac{7-1}{2} = 3\\x_2=\frac{7+1}{2} = 4[/tex]
Reasumując:
x ∈ (-∞, 3) ∪ (4, +∞)
Pozdrawiam.
