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