7. Given $1 \leqslant x^{2}+y^{2} \leqslant 4$, then the sum of the maximum and minimum values of $x^{2}-x y+y^{2}$ is

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7. $6 \frac{1}{2}$

Let $x=p \sin \theta, y=p \cos \theta, p>0, \theta \in[0,2 \pi)$, then $1 \leqslant p \leqslant 2, x^{2}-x y+y^{2}=p^{2}\left(1-\frac{1}{2} \sin 2 \theta\right)$. It is easy to find that the maximum and minimum values are $6$ and $\frac{1}{2}$, respectively.