7. Given the equation $x^{2}+(2-k) x+1=0$ satisfies the condition $x>-1$, and $k x>0$ has only one real root. Then the range of values for $k$ is $\qquad$ .
7. $k<0$ or $k=4$.

From the problem, we get $x_{1,2}=\frac{1}{2}\left(k-2 \pm \sqrt{k^{2}-4 k}\right)$.
Also, $\Delta=k^{2}-4 k \geqslant 0$, so $k \leqslant 0$ or $k \geqslant 4$.
By case analysis, we get $k<0$ or $k=4$.