小女不才,请各位大虾帮忙!求函数f（x）=(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x)

f（x）=(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x)
=((sin^2x+cos^2x)^2-sin^2xcos^2x)/(2-sin2x)
=(1-sin^2xcos^2x)/(2-sin2x)
=(1-sinx cosx)(1+sinxcosx)/(2-sin2x)
=(1-0.5xin2x)(1+0.5sin2x）/(2-sin2x)
=(1+0.5sin2x）/2=0.5+0.25sin2x
sin2x的最大值为1,最小值为-1
所以
y最大=0.75
y最小=0.25

解 :
f(x)
=[(sin^2 x+cos^2 x)^2-sin^2 x*cos^2 x]/(2-sin2x)
=(1-1/4 sin^2 2x)/(2-sin 2x)
设 t=sin 2x ,t∈[-1,1]
则f(x)可化为
g(t)=1/4*(t^2-4)/(t-2)=1/4*(t+2)
t=-1时，g(t)取min=...