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若tanα,tanβ是关于x的方程mx2-(2m-3)x+m-2=0的二个实根,求tan(α+β)的取值范围?
人气:437 ℃ 时间:2020-02-06 11:29:46
解答
tga,tgb为方程mx^2-(2m-3)x+m-2=0的两实根
则tga+tgb=(2m-3)/m,tga*tgb=(m-2)/m
tg(a+b)=(tga+tgb)/(1-tgatgb)
=((2m-3)/m)/(1-(m-2)/m)
=(2m-3)/2=m-3/2
方程有两实根
则判别式=(2m-3)^2-4m(m-2)=-4m+9>=0
则m
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