已知tan(x-π\4)=2,求(sin2x+2cos2x)/(2cosx平方-3sin2x-1)的值

tan(x-π/4)=2
=(tanx-tanπ/4)/(1+tanxtanπ/4)
=(tanx-1)/(1+tanx)
所以,tanx=-3
则tan2x=2tanx/(1-tan^2x)
=-6/(1-9)
=-6/-8
=3/4
(sin2x+2cos2x)/(2cos^2x-3sin2x-1)
=(sin2x+2cos2x)/(cos2x+1-3sin2x-1)
=(sin2x+2cos2x)/(cos2x-3sin2x)
=(tan2x+2)/(1-3tan2x)
=(2+3/4)/(1-3*3/4)
=(11/4)/(-5/4)
=-11/5
不懂的欢迎追问,如有帮助请采纳,谢谢!

tan(x-π4)=(tanx-1)/(1+tanx)=2 , tanx-1=2+2tanx , tanx=-3 ,tan2x=2tanx/(1-tan^2x)=3/4
(sin2x+2cos2x)/(2cos^2x-3sin2x-1)=(sin2x+2cos2x)/(cos2x-3sin2x)
=(tan2x+2)/(1-3tan2x)
=(3/4+2)/(1-9/4)
=-11/5