16. Let
$$
F(x)=\frac{1}{\sqrt{x+2 \sqrt{x-1}}}+\frac{1}{\sqrt{x-2 \sqrt{x-1}}} \text {, }
$$

then $F\left(\frac{3}{2}\right)$ is (
(A) 2
$\sqrt{3}$
(B) 3.5
(C) 4
(D) $3 \sqrt{2}$
(E) None of the above answers is correct
)
16. (C) Method 1 Notice that $x \pm 2 \sqrt{x-1}=1 \pm 2 \sqrt{x-1}+(x-1)=(1 \pm \sqrt{x-1})^{2}$, thus
$$
\begin{array}{l}
F\left(\frac{3}{2}\right)=\frac{1}{1+\sqrt{\frac{3}{2}-1}}+\frac{1}{1-\sqrt{\frac{3}{2}-1}} \\
=\frac{\sqrt{2}}{\sqrt{2}+1}+\frac{\sqrt{2}}{\sqrt{2}-1} \\
=\sqrt{2}(\sqrt{2}-1)+\sqrt{2}(\sqrt{2}+1)=4 .
\end{array}
$$

Method 2 Let $a(x)=x+2 \sqrt{x-1}$,
$$
b(x)=x-2 \sqrt{x-1}, 1 \leqslant x \leqslant 2 .
$$

Then, $a(x)+b(x)=2 x$,
and $a(x) \cdot b(x)=x^{2}-4(x-1)=(2-x)^{2}$,
$$
\begin{array}{l}
(\sqrt{a(x)}+\sqrt{b(x)})^{2} \\
=a(x)+b(x)+2 \sqrt{a(x) b(x)} \\
=2 x+2(2-x)=4 .
\end{array}
$$

Therefore, $F(x)=\frac{1}{\sqrt{a(x)}}+\frac{1}{\sqrt{b(x)}}=\frac{\sqrt{a(x)}+\sqrt{b(x)}}{\sqrt{a(x) b(x)}}$
$$
=\frac{2}{2-x} \text {. }
$$