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

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