5. (10 points) Define a new operation: $\theta_{a}=\frac{1}{\mathrm{a}^{2}-1}$, then $\left(\theta_{3}\right)+\left(\theta_{5}\right)+\left(\theta_{7}\right)\left(+\theta_{9}\right)+\left(\theta_{11}\right)$, after simplifying the result to the simplest proper fraction, the sum of the numerator and the denominator is
5. (10 points) Define a new operation: $\theta_{a}=\frac{1}{a^{2}-1}$, then the calculation result of $\left(\theta_{3}\right)+\left(\theta_{5}\right)+\left(\theta_{7}\right)\left(+\theta_{9}\right)+\left(\theta_{11}\right)$, when simplified to the simplest proper fraction, the sum of the numerator and the denominator is 29 $\qquad$
【Solution】Solution: The original expression $=\frac{1}{3^{2}-1}+\frac{1}{5^{2}-1}+\frac{1}{7^{2}-1}+\frac{1}{9^{2}-1}+\frac{1}{11^{2}-1}$
$$
\begin{array}{l}
=\frac{1}{2 \times 4}+\frac{1}{4 \times 6}+\frac{1}{6 \times 8}+\frac{1}{8 \times 10}+\frac{1}{10 \times 12} \\
=\frac{1}{2} \times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\cdots+\frac{1}{10}-\frac{1}{12}\right) \\
=\frac{1}{2} \times\left(\frac{1}{2}-\frac{1}{12}\right) \\
=\frac{5}{24} \\
5+24=29
\end{array}
$$

Therefore, the answer is: 29