\(\frac{x^2-2x+9}{2\left(x-3\right)\left(x+1\right)}\le0\) \(\left(x\ne3;-1\right)\)
\(\Leftrightarrow\frac{\left(x-1\right)^2+8}{2\left(x-1\right)^2-8}\le0\)
Mà \(\left(x-1\right)^2+8>0\) \(\forall x\ne-1;3\)
\(\Rightarrow2\left(x-1\right)^2-8< 0\Leftrightarrow2\left(x-1\right)^2< 8\)
\(\Leftrightarrow\left(x-1\right)^2< 4\Leftrightarrow-1< x< 3\)
Vậy \(-1< x< 3\)