65 If for any $a \in[-1,1]$, the inequality $x^{2}+(a-4) x+4-2 a>0$ holds, then the range of $x$ is $\qquad$ .
$$
65 x>3 \text{ or } x0
$$

and
$$
x-2+x^{2}-4 x+4>0
$$

Solving it, we get $x>3$ or $x3$, for $a \in[-1,1]$,
$$
x^{2}+(a-4) x+4-2 a>x^{2}-4 x+4-(x-2)>0
$$

For $xx^{2}-4 x+4+(x-2)>0 .
$$