作业帮已知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
来源：学生作业帮助网 编辑：作业帮 时间：2024/07/19 15:50:57
x){}KK*tDM|EwY&g<ٱق:OvL=ߠo~Ϝ
1u`0 3++6yc=
t:%t g0Y4(EO* $م`+Iu~PS6
r.AO
GvC Nyigs:hxcГ/M醍/:~y*Hng !19{. VXDN
^6-~(&jM0ԅ]1>}
]c
@X$]l{fa
L
Hu6<ٽ$ Myvd Z.
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