$33 \cdot 14$ In the set of real numbers, let $x$ be
$$
\left(\frac{\sqrt{(a-2)(|a-1|-1)}+\sqrt{(a-2)(1-|a|)}}{1+\frac{1}{1-a}}+\frac{5 a+1}{1-a}\right)^{1988},
$$

then the unit digit of $x$ is
(A) 1 .
(B) 2 .
(C) 4 .
(D) 6 .
(China Junior High School Mathematics League, 1988)
[Solution] For the two radicals to be meaningful, it is necessary that
$$
(a-2)(|a|-1) \geqslant 0 \text {, and } \quad(a-2)(1-|a|) \geqslant 0 \text {. }
$$

But \quad(a-2)(1-|a|)=-(a-2)(|a|-1),
so
$$
(a-2)(|a|-1)=0,
$$

Solving this gives \(a_{1}=2, a_{2}=1, a_{3}=-1\).
If \(a_{1}=2\), then \(1+\frac{1}{1-a}=0\);
If \(a_{2}=1\), then \(1-a=0\), both making the denominator zero.
Therefore, only \(a_{3}=-1\) is applicable. In this case,
$$
\begin{aligned}
x & =\left[\frac{5 \cdot(-1)+1}{1-(-1)}\right]^{1988} \\
& =(-2)^{1988} \\
& =2^{4 \cdot 497}=16^{497} .
\end{aligned}
$$

Thus, the units digit of \(x\) is 6 (or by noting that the last digit of \(2^n\) cycles every 4 as \(n\) changes).
Therefore, the answer is \((D)\).