求函数f(x)=（sin^4x+cos^4x+sin^2xcos^2x）/（2-sin2x）的最小正周期、最大值和最小值

f(x)=[(sin^2x+cos^2x)^2-sin^2xcos^2x]/(2-2sinxcosx)
=(1-sinxcosx)(1+sinxcosx)/2(1-sinxcosx)
=1/2sinxcosx+1/2
=1/4sin2x+1/2
∴T=2π/2=π
∴f(x)max=1/4+1/2=3/4
∴f(x)min=-1/4+1/2=1/4