12. Let $f(x)=x^{2}-(k+1) x+2$. If for $x>0$, $f(x)$ is always greater than zero, then the range of values for $k$ is $\qquad$.
12. $(-\infty, 2 \sqrt{2}-1)$.

From $x^{2}-(k+1) x+2>0 \Rightarrow k+1<x+\frac{2}{x}$.
Also, $x+\frac{2}{x} \geqslant 2 \sqrt{2}$, equality holds when $x=\sqrt{2}$, at this point, $k<2 \sqrt{2}-1$.