4. $45 f(x)=x^{3}-3 x+\left(x^{2}-1\right) \sqrt{x^{2}-4}$, Express $\frac{f(x)-2}{f(x)+2}$ as a fraction where both the numerator and the denominator are in the form $u \sqrt{v}$, where $u, v$ are linear polynomials in $x$, and state for which values of $x$ your equation holds.
[Solution] If $x \geqslant 2$, then
$$\begin{aligned}
\frac{f(x)-2}{f(x)+2} & =\frac{x^{3}-3 x-2+\left(x^{2}-1\right) \sqrt{x^{2}-4}}{x^{3}-3 x+2+\left(x^{2}-1\right) \sqrt{x^{2}-4}} \\
& =\frac{(x+1)^{2}(x-2)+(x+1)(x-1) \sqrt{x^{2}-4}}{(x-1)^{2}(x+2)+(x+1)(x-1) \sqrt{x^{2}-4}} \\
& =\frac{(x+1) \sqrt{x-2}[(x+1) \sqrt{x-2}+(x-1) \sqrt{x+2}]}{(x-1) \sqrt{x+2}[(x-1) \sqrt{x+2}+(x+1) \sqrt{x-2}]} \\
& =\frac{(x+1) \sqrt{x-2}}{(x-1) \sqrt{x+2}} .
\end{aligned}$$

If $x \leqslant-2$, then
$$\begin{aligned}
& \frac{f(x)-2}{f(x)+2} \\
= & \frac{-(x+1) \sqrt{-x+2}[(x+1) \sqrt{-x+2}-(x-1) \sqrt{-x-2}]}{(x-1) \sqrt{-x-2}[-(x-1) \sqrt{-x-2}+(x+1) \sqrt{-x+2}]} \\
= & \frac{-(x+1) \sqrt{-x+2}}{(x-1) \sqrt{-x-2}} .
\end{aligned}$$