1. Given $m^{2}+n^{2}+m n+m-n+1=0$. Then the value of $\frac{1}{m}+\frac{1}{n}$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 2
$-1 . B$.
Completing the square gives $-\frac{1}{2}\left[(m+n)^{2}+(m+1)^{2}+(n-1)^{2}\right]=0$.
Therefore, $m=-1, n=1$. Hence $-\frac{1}{m}+\frac{1}{n}=0$.