5. Given that $x, y$ are in the interval $(-2,2)$, and $x y=-1$, then the minimum value of the function $u=\frac{4}{4-x^{2}}+\frac{9}{9-y^{2}}$ is
(A) $\frac{8}{5}$
(B) $\frac{24}{11}$
(C) $\frac{12}{7}$
(D) $\frac{12}{5}$
(D)
5.【Analysis and Solution】From the given, we have $y=-\frac{1}{x}$, thus
$$
\begin{aligned}
u & =\frac{4}{4-x^{2}}+\frac{9 x^{2}}{9 x^{2}-1}=\frac{-9 x^{4}+72 x^{2}-4}{-9 x^{4}+37 x^{2}-4} \\
& =1+\frac{35}{37-\left(9 x^{2}+\frac{4}{x^{2}}\right)}
\end{aligned}
$$

Given $x \in\left(-2,-\frac{1}{2}\right) \cup\left(\frac{1}{2}, 2\right)$, when $9 x^{2}=\frac{4}{x^{2}}$, i.e., $x^{2}=\frac{2}{3}$, the value of $9 x^{2}+\left(\frac{4}{x^{2}}\right)$ is minimized, and at this point, the function $u$ has a minimum value of $\frac{12}{5}$. Therefore, the answer is (D).