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_数学解答

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

人气：492 ℃　时间：2020-09-08 12:15:52

解答

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...