7. If real numbers $x, y$ satisfy $2 x^{2}+3 x y+2 y^{2}=1$, then the minimum value of $x+y+x y$ is $\qquad$ .
7. $-\frac{9}{8}$.

Let $k=x+y+xy$.
From $2x^{2}+3xy+2y^{2}=1 \Rightarrow 2(x+y)^{2}=1+xy$
$\Rightarrow 2(x+y)^{2}=k-(x+y)+1$
$\Rightarrow 2(x+y)^{2}+(x+y)-(1+k)=0$.
Consider the above equation as a quadratic equation in terms of $x+y$.
Since $x+y$ is a real number, we have
$$
\Delta=1+8(1+k) \geqslant 0 \Rightarrow k \geqslant-\frac{9}{8} \text {. }
$$

Therefore, the minimum value of $x+y+xy$ is $-\frac{9}{8}$.