$2 \cdot 120$ If $a b \neq 0,|a| \neq|b|$, then the number of different $x$ that satisfy the equation
$$
\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b}
$$

is
(A) 0 .
(B) 1 .
(C) 2 .
(D) 3 .
(E) 4 .
(17th American High School Mathematics Examination, 1966)
[Solution] The given equation can be transformed into
$$
\frac{a(x-a)+b(x-b)}{a b}=\frac{b(x-b)+a(x-a)}{(x-a)(x-b)}
$$

The numerators are the same, so if the fractions are equal, either the numerator is 0, or the denominators are equal.
(1) If the numerator is 0, then
$a(x-a)+b(x-b)=0$, which gives $x=\frac{a^{2}+b^{2}}{a+b}$;
(2) If the denominators are equal, then
$(x-a)(x-b) \doteq a b$, solving this yields $\quad x=0$ or $x=a+b$. Therefore, the answer is $(D)$.