1. Given $a+b+c=0$, let
$$
\begin{array}{l}
M=\frac{a^{2}}{2 a^{2}+b c}+\frac{b^{2}}{2 b^{2}+a c}+\frac{c^{2}}{2 c^{2}+a b}, \\
N=\frac{b c}{2 a^{2}+b c}+\frac{a c}{2 b^{2}+a c}+\frac{a b}{2 c^{2}+a b} .
\end{array}
$$

Then the relationship between $M$ and $N$ is ().
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) Cannot be determined
-1. B.
Given $a+b+c=0$, we have
$$
\begin{array}{l}
2 a^{2}+b c=2 a^{2}+b(-a-b) \\
=a^{2}-a b+a^{2}-b^{2}=(a-b)(a-c) .
\end{array}
$$

Similarly, $2 b^{2}+a c=(b-c)(b-a)$,
$$
2 c^{2}+a b=(c-a)(c-b) \text {. }
$$

Thus, $M=-\frac{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)}{(a-b)(b-c)(c-a)}$.
And $N=\frac{b c}{(a-b)(a-c)}+\frac{a c}{(b-a)(b-c)}+$
$$
\begin{array}{l} 
\frac{a b}{(c-a)(c-b)} \\
=-\frac{b c(b-c)+a c(c-a)+a b(a-b)}{(a-b)(b-c)(c-a)} \\
=-\frac{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)}{(a-b)(b-c)(c-a)} .
\end{array}
$$

Therefore, $M=N$.