1. Given real numbers $x, y$ satisfy $\left(x^{2}+6 x+12\right)\left(5 y^{2}+2 y+1\right)=\frac{12}{5}$. Then the value of $x y$ is $\qquad$ (2011, Huanggang City, Hubei Province Mathematics Competition)
$\begin{array}{l}\left.12=\left((x+3)^{2}+3\right)\right)\left((5 y+1)^{2}+4\right) \geqslant 12 \\ \Rightarrow x=-3, y=-\frac{1}{5} \Rightarrow x y=\frac{3}{5} .\end{array}$

The translation is as follows:

$\begin{array}{l}\left.12=\left((x+3)^{2}+3\right)\right)\left((5 y+1)^{2}+4\right) \geqslant 12 \\ \Rightarrow x=-3, y=-\frac{1}{5} \Rightarrow x y=\frac{3}{5} .\end{array}$