1. Given the quadratic equation in $x$
$$
(n+1)(n+2) x^{2}+x-n(n+3)=0
$$

with two real roots $a_{n}$ and $b_{n}\left(n \in \mathbf{N}_{+}\right)$. Then the value of $a_{1} a_{2} \cdots a_{2000} \cdot b_{1} b_{2} \cdots b_{2000}$ is $\qquad$
$$
=1 -\frac{1006}{3015} \text {. }
$$

According to Vieta's formulas, we have
$$
\begin{array}{l}
a_{n} b_{n}=-\frac{n(n+3)}{(n+1)(n+2)} . \\
\text { Therefore, } a_{1} a_{2} \cdots a_{2009} \cdot b_{1} b_{2} \cdots b_{2009} \\
=a_{1} b_{1} \cdot a_{2} b_{2} \cdots \cdot a_{2009} b_{2009} \\
=\left(-\frac{1 \times 4}{2 \times 3}\right)\left(-\frac{2 \times 5}{3 \times 4}\right) \cdots\left(-\frac{2009 \times 2012}{2010 \times 2011}\right) \\
=-\frac{(1 \times 2 \times \cdots \times 2009)(4 \times 5 \times \cdots \times 2012)}{(2 \times 3 \times \cdots \times 2010)(3 \times 4 \times \cdots \times 2011)} \\
=-\frac{1 \times 2012}{2010 \times 3}=-\frac{1006}{3015} .
\end{array}
$$