9. 56 Find
$$\begin{aligned}
A & =\sqrt{\left(1264-z_{1}-\cdots-z_{n}\right)^{2}+x_{n}^{2}+y_{n}^{2}}+ \\
& \sqrt{z_{n}^{2}+x_{n-1}^{2}+y_{n-1}^{2}}+\cdots+\sqrt{z_{2}^{2}+x_{1}^{2}+y_{1}^{2}}+ \\
& \sqrt{z_{1}^{2}+\left(948-x_{1}-\cdots-x_{n}\right)^{2}+\left(1185-y_{1}-\cdots-y_{n}\right)^{2}}
\end{aligned}$$

the minimum value, where $x_{i}, y_{i}, z_{i}, i=1,2, \cdots, n$, are non-negative real numbers. Find: $M=\min _{\alpha, \beta} \max _{x} f(x)$.
[Solution] Consider the following points in a three-dimensional space with a Cartesian coordinate system:
$$\begin{array}{l}
B=(0,0,1264) \\
M_{n}=\left(x_{n}, y_{n}, z_{1}+\cdots+z_{n}\right) \\
M_{n-1}=\left(x_{n-1}+x_{n}, y_{n-1}+y_{n}, z_{1}+\cdots+z_{n-1}\right) \\
\cdots \cdots \cdots \cdots \\
M_{2}=\left(x_{2}+\cdots+x_{n}, y_{2}+\cdots+y_{n}, z_{1}+z_{2}\right) \\
M_{1}=\left(x_{1}+\cdots+x_{n}, y_{1}+\cdots y_{n}, z_{1}\right) \\
C=(948,1185,0)
\end{array}$$

Clearly, $A=B M_{n}+M_{n} M_{n-1}+\cdots+M_{2} M_{1}+M_{1} C$. By the triangle inequality, we have $A \geqslant B C=\sqrt{948^{2}+1185^{2}+1264^{2}}=1975$. On the other hand, if we take $M_{1}=M_{2}=\cdots=M_{n}=B$, then $A=B C=1975$. Therefore, the minimum value of $A$ is 1975.
9・57 Let the function $f(x)=|\cos x+\alpha \cos 2 x+\beta \cos 3 x|$, where $\alpha, \beta$ are
[Solution] Clearly, $f\left(\frac{\pi}{6}\right)=\left|\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|, f\left(\frac{5}{6} \pi\right)=\left|-\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|$, 
thus $\max f(x) \geqslant \frac{1}{2}\left(\left|\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|+\left|-\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|\right)$
$$\geqslant \frac{1}{2}\left[\left(\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right)-\left(-\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right)\right]=\frac{\sqrt{3}}{2}$$

Thus, we have $\quad M \geqslant \frac{\sqrt{3}}{2}$.
On the other hand, let $\alpha=0, \beta=-\frac{1}{6}$, then
$f(x)=\left|\cos x-\frac{1}{6} \cos 3 x\right|=\left|\frac{3}{2} \cos x-\frac{2}{3} \cos ^{3} x\right|$. It is easy to see that
$$\max _{x} f(x)=\max _{-1 \leqslant y \leqslant 1}|g(y)|=\max _{0 \leqslant y \leqslant 1} g(y),$$

where $g(y)=\frac{3}{2} y-\frac{2}{3} y^{3}$. Since
$$g(y)-g\left(\frac{\sqrt{3}}{2}\right)=\left(y-\frac{\sqrt{3}}{2}\right)\left[\frac{3}{2}-\frac{2}{3}\left(y^{2}+\frac{\sqrt{3}}{2} y+\frac{3}{4}\right)\right]$$

Therefore, when $0 \leqslant y \leqslant \frac{\sqrt{3}}{2}$, by $y^{2}+\frac{\sqrt{3}}{2} y+\frac{3}{4} \leqslant \frac{9}{4}$ we get
$$g(y)-g\left(\frac{\sqrt{3}}{2}\right) \leqslant 0$$

When $\frac{\sqrt{3}}{2} \leqslant y \leqslant 1$, by $y^{2}+\frac{\sqrt{3}}{2} y+\frac{3}{4} \geqslant \frac{9}{4}$ we get
$$g(y)-g\left(\frac{\sqrt{3}}{2}\right) \leqslant 0$$

Thus, we have $\quad \max _{x} f(x)=\max _{0 \leqslant y \leqslant 1} g(y)=g\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{2}$.
Therefore, we have $\quad M \leqslant \frac{\sqrt{3}}{2}$.
Combining (1) and (2), we get $M=\frac{\sqrt{3}}{2}$.