9. Let $P$ be any point in the first quadrant on the hyperbola $\frac{x^{2}}{3}-y^{2}=1$, and let $F_{1}$ and $F_{2}$ be the left and right foci, respectively. The lines $P F_{1}$ and $P F_{2}$ intersect the hyperbola at points $M$ and $N$, respectively. If $P F_{1}=\lambda_{1} F_{1} M$ and $P F_{2}=\lambda_{2} F_{2} N$, find the value of $\lambda_{1}+\lambda_{2}$ and the range of the slope $k_{M N}$ of the line $M N$.
9. Analysis: Let the coordinates of point $P$ be $\left(x_{0}, y_{0}\right)$, $M$ be $\left(x_{1}, y_{1}\right)$, and $N$ be $\left(x_{2}, y_{2}\right)$. The equation of the line $P F_{1}$ is: $y=\frac{y_{0}}{x_{0}+2} \cdot(x+2)$. By combining the hyperbola equation and the line equation, we get:
$$
\left(1-\frac{3 y_{0}^{2}}{\left(x_{0}+2\right)^{2}}\right) x^{2}-\frac{12 y_{0}^{2}}{\left(x_{0}+2\right)^{2}} \cdot x-\frac{12 y_{0}^{2}}{\left(x_{0}+2\right)^{2}}-3=0
$$

By Vieta's formulas: $x_{0}+x_{1}=\frac{12 y_{0}^{2}}{\left(x_{0}+2\right)^{2}-3 y_{0}^{2}}$, similarly, we get: $x_{0}+x_{2}=-\frac{12 y_{0}^{2}}{\left(x_{0}-2\right)^{2}-3 y_{0}^{2}}$
Thus, $x_{1}=\frac{12 y_{0}^{2}}{\left(x_{0}+2\right)^{2}-3 y_{0}^{2}}-x_{0}=-\frac{12+7 x_{0}}{4 x_{0}+7}, \quad x_{2}=-\frac{12 y_{0}^{2}}{\left(x_{0}-2\right)^{2}-3 y_{0}^{2}}-x_{0}=\frac{7 x_{0}-12}{4 x_{0}-7}$
Then $\lambda_{1}+\lambda_{2}=-\left(\frac{2+x_{0}}{2+x_{1}}+\frac{2-x_{0}}{2-x_{2}}\right)=-\left(\frac{2+x_{0}}{2-\frac{12+7 x_{0}}{4 x_{0}+7}}+\frac{2-x_{0}}{2-\frac{7 x_{0}-12}{4 x_{0}-7}}\right)=-14$ is a constant
$$
k_{M N}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{\frac{y_{0}}{x_{0}+2} \cdot\left(x_{1}+2\right)-\frac{y_{0}}{x_{0}-2} \cdot\left(x_{2}-2\right)}{x_{1}-x_{2}}=-\frac{\frac{y_{0}}{4 x_{0}+7}+\frac{y_{0}}{4 x_{0}-7}}{\frac{12+7 x_{0}}{4 x_{0}+7}+\frac{7 x_{0}-12}{4 x_{0}-7}}=-\frac{x_{0} y_{0}}{7 x_{0}^{2}-21}=-\frac{x_{0}}{21 y_{0}}
$$

Since $P$ is on the hyperbola, $\frac{x_{0}}{y_{0}} \in(\sqrt{3},+\infty)$, hence $k_{M N} \in\left(-\infty,-\frac{\sqrt{3}}{21}\right)$