7. If the equation $\frac{|x|}{x+6}=k x^{2}$ has four distinct real solutions for $x$, then the range of real values for $k$ is $\qquad$.

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7. $\left(\frac{1}{9},+\infty\right)$.

Obviously, $x=0$ is a solution to the original equation.
If $x \neq 0$, then
$$
k=\frac{1}{|x|(x+6)}=\left\{\begin{array}{ll}
\frac{1}{x(x+6)}, & x>0 ; \\
\frac{1}{-x(x+6)}, & x>0 ; \\
-x(x+6), & x<0
\end{array}\right.
$$

There are three non-zero solutions.
As shown in Figure 3, we know $0<\frac{1}{k}<9$, i.e., $k \in\left(\frac{1}{9},+\infty\right)$.
Also, $k \neq 0$, hence
$$
k \in\left(\frac{1}{9},+\infty\right) .
$$