［Solution］By Vieta's formulas, we get $a^{2}+b^{2}=c, a^{2} b^{2}=c$, hence $\quad a^{2}+b^{2}=a^{2} b^{2}, \quad \therefore \quad b^{2}-1=\frac{b^{2}}{a^{2}}, \quad a^{2}-1=\frac{a^{2}}{b^{2}}$. If $a>0, b>0$, then
$$
\begin{aligned}
& a \sqrt{1-\frac{1}{b^{2}}}+b \sqrt{1-\frac{1}{a^{2}}} \\
= & \sqrt{a^{2}-\frac{a^{2}}{b^{2}}}+\sqrt{b^{2}-\frac{b^{2}}{a^{2}}} \\
= & \sqrt{a^{2}-\left(a^{2}-1\right)}+\sqrt{b^{2}-\left(b^{2}-1\right)}=2 .
\end{aligned}
$$

If $a<0, b<0$, the original expression $=0$. Therefore, the answer is $(D)$.
2.71 $a, b$ are two non-zero real numbers, $a^{2}, b^{2}$ are the roots of the equation $x^{2}-c x+c=0$, then the value of the algebraic expression $a \sqrt{1-\frac{1}{b^{2}}}+b \sqrt{1-\frac{1}{a^{2}}}$ is
(A) 0.
(B) 2.
(C) Does not exist.
(D) None of the above answers is correct.
(China Jiangxi Province Nanchang City Junior High School Mathematics Competition, 1990)