9. Given that $\alpha, \beta$ are the two roots of the quadratic equation $2 x^{2}-t x-2=0$ with respect to $x$, and $\alpha<\beta$, if the function $f(x)=$ $\frac{4 x-t}{x^{2}+1}$.
(1) Find the value of $\frac{f(\alpha)-f(\beta)}{\alpha-\beta}$;
(2) For any positive numbers $\lambda_{1}, \lambda_{2}$, prove that $\left|f\left(\frac{\lambda_{1} \alpha+\lambda_{2} \beta}{\lambda_{1}+\lambda_{2}}\right)-f\left(\frac{\lambda_{1} \beta+\lambda_{2} \alpha}{\lambda_{1}+\lambda_{2}}\right)\right|<2|\alpha-\beta|$.
9. (1) By Vieta's formulas, we have $\alpha+\beta=\frac{t}{2}, \alpha \cdot \beta=-1, f(\alpha)=\frac{4 \alpha-t}{\alpha^{2}+1}=\frac{4 \alpha-2(\alpha+\beta)}{\alpha^{2}-\alpha \beta}=\frac{2}{\alpha}=-2 \beta$,
$$
f(\beta)=\frac{4 \beta-t}{\beta^{2}+1}=\frac{4 \beta-2(\alpha+\beta)}{\beta^{2}-\alpha \beta}=\frac{2}{\beta}=-2 \alpha, \therefore \frac{f(\alpha)-f(\beta)}{\alpha-\beta}=\frac{-2 \beta+2 \alpha}{\alpha-\beta}=2 .
$$
(2) Given the function $f(x)=\frac{4 x-t}{x^{2}+1}, \therefore f^{\prime}(x)=\frac{-2\left(2 x^{2}-t x-2\right)}{\left(x^{2}+1\right)^{2}}$.

Moreover, for $x \in[\alpha, \beta], 2 x^{2}-t x-2=2(x-\alpha)(x-\beta) \leqslant 0$, thus $f^{\prime}(x) \geqslant 0$,
$\therefore$ the function $f(x)=\frac{4 x-t}{x^{2}+1}$ is increasing on $[\alpha, \beta]$. Noting that for any positive numbers $x_{1} 、 x_{2}$ we have
$$
\frac{x_{1} \alpha+x_{2} \beta}{x_{1}+x_{2}}-\alpha=\frac{x_{2}(\beta-\alpha)}{x_{1}+x_{2}}>0, \frac{x_{1} \alpha+x_{2} \beta}{x_{1}+x_{2}}-\beta=\frac{x_{1}(\alpha-\beta)}{x_{1}+x_{2}}<0,
$$

which means $\alpha<\frac{x_{1} \alpha+x_{2} \beta}{x_{1}+x_{2}}<\beta$, similarly $\alpha<\frac{x_{1} \beta+x_{2} \alpha}{x_{1}+x_{2}}<\beta$.
$$
\begin{array}{l}
\therefore f(\alpha)<f\left(\frac{x_{1} \alpha+x_{2} \beta}{x_{1}+x_{2}}\right)<f(\beta), f(\alpha)<f\left(\frac{x_{1} \beta+x_{2} \alpha}{x_{1}+x_{2}}\right)<f(\beta), \\
-f(\beta)<-f\left(\frac{x_{1} \beta+x_{2} \alpha}{x_{1}+x_{2}}\right)<-f(\alpha) .
\end{array}
$$

Thus, $-[f(\beta)-f(\alpha)]<f\left(\frac{x_{1} \alpha+x_{2} \beta}{x_{1}+x_{2}}\right)-f\left(\frac{x_{1} \beta+x_{2} \alpha}{x_{1}+x_{2}}\right)<f(\beta)-f(\alpha)$,
$\therefore\left|f\left(\frac{x_{1} \alpha+x_{2} \beta}{x_{1}+x_{2}}\right)-f\left(\frac{x_{1} \beta+x_{2} \alpha}{x_{1}+x_{2}}\right)\right|<f(\beta)-f(\alpha)$. And $f(\beta)-f(\alpha)=2 \beta-2 \alpha=2|\alpha-\beta|$,
$$
\therefore\left|f\left(\frac{x_{1} \alpha+x_{2} \beta}{x_{1}+x_{2}}\right)-f\left(\frac{x_{1} \beta+x_{2} \alpha}{x_{1}+x_{2}}\right)\right|<2|\alpha-\beta| .
$$