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

显示全部楼层 · 2013-4-8 23:08:33

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=<9/4 所以tan(a +b)的最小值是(3-9/2)/2=-3/4...

千问 · 2013-4-8 23:08:33

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