y=(sinxcosx-1)/(sinx+cosx+1)的最值问题
网友回答
设sinx+cosx=t
sinxcosx=(t^2-1)/2
y=[(t^2-1)/2]/(t+1)
=(t+1)(t-1)/[2(t+1)]
=(t-1)/2
t=sinx+cosx=√2sin(x+π/4)∈(-1,1]
∴ymax=f(1)=(1-1)/2=0
======以下答案可供参考======
供参考答案1:
sinx+cosx=t,-√2y=(sinxcosx-1)/(sinx+cosx+1)
=[1/2(sinx+cosx)^2-3/2]/(sinx+cosx+1)
=1/2*[t^2-3)/(t+1),t≠-1
y≠0t^2-2yt-2y-3=0
判别4y^2-4*(-2y-3)
(y+1)^2+2>0,所以：y0 没有最值