7. $\frac{3+k_{1}+3 k_{2}}{1-k_{1}-k_{2}}$.

By Vieta's formulas, we know
$$
\begin{array}{l}
\sum a=0, \sum a b=-k_{1}, a b c=k_{2} . \\
\text { Therefore, } \frac{1+a}{1-a}=\frac{2 \sum(1-b)(1-c)}{\prod(1-a)}-3 \\
=\frac{2\left(3+\sum b c\right)}{1+\sum a b-a b c}-3=\frac{3+k_{1}+3 k_{2}}{1-k_{1}-k_{2}} .
\end{array}
$$
7. Let $a$, $b$, $c$ be the roots of the equation
$$
x^{3}-k_{1} x-k_{2}=0\left(k_{1}+k_{2} \neq 1\right)
$$

Then
$$
\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}=
$$
$\qquad$