30. Solve the system of equations
$$
\begin{array}{l}
2 y=x+\frac{17}{x}, 2 z=y+\frac{17}{y} \quad 2 w=z+\frac{17}{z}, \\
2 x=w+\frac{17}{w}
\end{array}
$$

The number of real solutions $(x, y, z, w)$ is
(A) 1 ; (B) 2 ; (C) 4 ; (D) 8 ; (E) 16 .
30
B
30. First, either $x>1$, or $x<0$, consider each equation in sequence, we can deduce that $y \geqslant \sqrt{17}$, $z \geqslant \sqrt{17}$, $w \geqslant \sqrt{17}$, and $x \geqslant \sqrt{17}$. Assume $x>\sqrt{17}$, then
$$
y-\sqrt{17}=\frac{x^{2}+17}{2 x}-\sqrt{17}=\left(\frac{x-\sqrt{17}}{2 x}\right)
$$
$\cdot(x-\sqrt{17})y$. Similarly, we get $y>z, z>w$ and $w>x$, thus leading to the obvious contradiction $x>x$.
Therefore, $x=y=z=w=\sqrt{17}$ is a solution, and indeed the only solution when $x>C$.

For $x<0$, note that $(x, y, z, w)$ is a solution if and only if $(-x, -y, -z, -w)$ is a solution. Therefore, $a=y=z=w=-\sqrt{17}$ is the only solution in this case.
(Beijing Mathematical Society整理, Liu Xihan, Zhou Chunli)