1. When $x$ varies, the minimum value of the fraction $\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$ is $\qquad$ .
1. 4
$$
\begin{aligned}
\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1} & =\frac{6 x^{2}+12 x+10}{x^{2}+2 x+2}=6-\frac{2}{x^{2}+2 x+2} \\
& =6-\frac{2}{(x+1)^{2}+1},
\end{aligned}
$$
$\therefore$ When $x=-1$, the formula takes the minimum value 4.