令t=sinx-cosx =√2[sinxcos(pi/4-cosxsin(pi/4)] =√2sin(x-pi/4) 0=-pi/4=-√2/2=-1=sin2x=1-t^2 所以y=t+1-t^2=-t^2+t+1=-(t-1/2)^2+5/4 -1=-3/2=0=-9/4=-1=<-(t-1/2)^2+5/4=<5/4 所以函数的值域是[-1,5/4]
答：解： 分母≠0 设t=sinx+cosx≠-1 sin2x=t²-1 f(t)=(t²-1)/(1+t)=t-1≠-2 又sinx+cosx...详情>>
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