若方程mx^2+(2m-3)x+m-2=0(m不等于0)的两根分别为tanA,tanB,求tan(A+B)的值

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/06 20:45:15
若方程mx^2+(2m-3)x+m-2=0(m不等于0)的两根分别为tanA,tanB,求tan(A+B)的值

若方程mx^2+(2m-3)x+m-2=0(m不等于0)的两根分别为tanA,tanB,求tan(A+B)的值
若方程mx^2+(2m-3)x+m-2=0(m不等于0)的两根分别为tanA,tanB,求tan(A+B)的值

若方程mx^2+(2m-3)x+m-2=0(m不等于0)的两根分别为tanA,tanB,求tan(A+B)的值
tana*tanb=(m-2)/m ,tana+tanb=(3-2m)/m
tan(a +b)=(tana+tanb)/(1-tana*tanb)
=(3-2m)/(m-(m-2)) =(3-2m)/2
又因为方程mx² + (2m - 3)x +(m - 2) = 0 (m¹0)有两个实根
(2m-3)^2-4m(m-2)>=0
4m^2-12m+9-4m^2+8m>=0
m=

∵tanA、tanB是方程mx^2+(2m-3)x+m-2=0的两个根,
∴tanA+tanB=-(2m-3)/m
tanA*tanB=(m-2)/m
∴tan(A+B)=(tanA+tanB)/(1-tanAtanB)
=[-(2m-3)/m]/[1-(m-2)/m]
=(3-2m)/2