$2 \cdot 30$ Given the equation $(x-m)^{2}-(x-n)^{2}=(m-n)^{2}, m$ is a positive number, $n$ is a negative number, then the set of $x$ values that satisfy the equation is
(A) $x \geqslant 0$.
(B) $x \leqslant n$.
(C) $x=0$.
(D) the set of all real numbers.
(E) none of the above.
(10th American High School Mathematics Examination, 1959)
［Solution］From the given equation, it can be transformed into

that is
$$
\begin{array}{c}
(x-m+x-n)(x-m-x+n)=(m-n)^{2}, \\
(2 x-m-n)(n-m)=(m-n)^{2} .
\end{array}
$$

Given $m>0, n<0$, hence $m-n \neq 0$. Dividing both sides of the above equation by $n-m$ yields $2 x-m-n=n-m$, which simplifies to $\quad x=n$. Therefore, the answer is $(E)$.