1.78 If $\sqrt{a}+\sqrt{b}=1$, and $\sqrt{a}=m+\frac{a-b}{2}, \sqrt{b}=n-\frac{a-b}{2}$, where $m, n$ are rational numbers, then
(A) $m n=\frac{1}{2}$.
(B) $m^{2}+n^{2}=\frac{1}{2}$.
(C) $m+n=\frac{1}{2}$.
(D) $m-n=\frac{1}{2}$.
(China "Zu Chongzhi Cup" Junior High School Mathematics Invitational Competition, 1993)
[Solution] From the given, we know
$$
\begin{aligned}
\sqrt{a} & =\frac{\sqrt{a}+\sqrt{b}}{2}+\frac{\sqrt{a}-\sqrt{b}}{2}=\frac{1}{2}+\frac{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}{2(\sqrt{a}+\sqrt{b})} \\
& =\frac{1}{2}+\frac{a-b}{2},
\end{aligned}
$$

Similarly, $\sqrt{b}=\frac{1}{2}-\frac{a-b}{2}$.
Thus, $m=\frac{1}{2}, n=\frac{1}{2}$, and therefore $m^{2}+n^{2}=\frac{1}{2}$. Hence, the answer is $(B)$.