1. Let $a<b<0, a^{2}+b^{2}=4 a b$. Then the value of $\frac{a+b}{a-b}$ is ( ).
(A) $\sqrt{2}$
(B) $\sqrt{3}$
(C) 2
(D) $\sqrt{6}$
-1. B.
From the given information, we have
$$
(a+b)^{2}=6 a b,(a-b)^{2}=2 a b \text {. }
$$

Since $a<b<0$, then,
$$
\frac{a+b}{a-b}=\sqrt{\frac{(a+b)^{2}}{(a-b)^{2}}}=\sqrt{\frac{6 a b}{2 a b}}=\sqrt{3} \text {. }
$$