已知tanx,tany是方程x²+3√3+4=0的两根,且-π/2＜x＜π/2,-π/2＜y＜π/2,则x+y的值为（ ）A.π/3 B.-2π/3 C.π/3或-2π/3 D.-π/3或2π/3

已知tanx,tany是方程x²+3√3+4=0的两根,且-π/2＜x＜π/2,-π/2＜y＜π/2,则x+y的值为（ ）A.π/3 B.-2π/3 C.π/3或-2π/3 D.-π/3或2π/3
已知tanx,tany是方程x²+3√3+4=0的两根,且-π/2＜x＜π/2,-π/2＜y＜π/2,则x+y的值为（ ）
A.π/3 B.-2π/3 C.π/3或-2π/3 D.-π/3或2π/3

已知tanx,tany是方程x²+3√3+4=0的两根,且-π/2＜x＜π/2,-π/2＜y＜π/2,则x+y的值为（ ）A.π/3 B.-2π/3 C.π/3或-2π/3 D.-π/3或2π/3
由tanx+tany= -3√30可知
-π/2＜x＜0,-π/2＜y＜0
答案只有一个是负的.就选B
方法是这样的
因为tanx+tany= -3√3,tanxtany=4,
那么tan(x+y)
=(tanx+tany)/(1-tanxtany)
= -3√3/(-3)
=√3
此时：x+y=π/3+kπ
所以x+y=-2π/3