1. Given $x$ and $y$ are real numbers, and $x^{2}+x y+y^{2}=3$. Let the maximum and minimum values of $x^{2}-x y+y^{2}$ be $m$ and $n$, respectively. Then the value of $m+n$ is $\qquad$
$=, 1.10$.
Given the equation is symmetric about $x, y$, let $x=a+b, y=a-b$, then $3=x^{2}+xy+y^{2}=3a^{2}+b^{2}$.
Therefore, $3a^{2}=3-b^{2}$.
So, $0 \leqslant b^{2} \leqslant 3$.
$$
\begin{array}{l}
\text { Also, } x^{2}-xy+y^{2}=a^{2}+3b^{2}=\frac{1}{3}\left(3a^{2}+b^{2}\right)+\frac{8}{3}b^{2} \\
=\frac{1}{3} \times 3+\frac{8}{3}b^{2}=1+\frac{8}{3}b^{2},
\end{array}
$$

Then $1 \leqslant 1+\frac{8}{3}b^{2} \leqslant 1+\frac{8}{3} \times 3=9$.
Thus, $m=9, n=1$.