1.f(x)=x^(1/-n^2+2n+3)=x^(-1/(n-3)(n+1)) 因为n＝2k，k∈Z,所以(n-3)(n+1)为奇数所以f(x)为奇函数.在〔0,＋∞〕上单调递增,那么在(-∞,0)也单调增,即在R里单调增f（x^2-x)>f(x+3),则x^2-x>x+3,解得x>3,或x<-12.f(16)=f(4*4)=f(4)+f(4)=4,得f(4)=2f(4)=f(2*2)=2f(2)=2,得f(2)=1f(x)在(0,＋∞〕上是单调递增函数f(x^2-2x+1.5)>1=f(2)所以x^2-2x+1.5>2解得x<1-(√6/2)或x>1+(√6/2) 3.f(x)为奇函数,且在[0,2]上单调递减,那么在[-2,0]也单调减即在[-2,2]单调减f(3-m)≤f(2m^2)3-m≥2m^2-2≤3-m≤20≤2m^2≤2解得:m=1

1. ∵ f(x)在〔0,＋∞〕上单调递增, 且f(x²-x)>f(x+3)∴ x²-x>0且x+3>0且x²-x>x+3,解得-3<x<-1或x>32. ∵ f(4*4)=2f(4)=4, ∴ f(4)=2, 同理,得f(2)=1而f(x)在(0,＋∞〕上是单调递增函数, 而f(x²-2x+1.5)>f(2)∴ x²-2x+1.5>2,x²-2x-0.5>0,解得x<1-(√6/2)或x>1+(√6/2)3. ∵ 在[-2,2]上的奇函数f(x)在[0,2]上单调递减,∴ f(x)在[-2,2]上单调递减.∴ -2≤3-m≤2且0≤2m²≤2且3-m≥2m²--->1≤m≤5且-1≤m≤1且-3/2≤m≤1, ∴ m=1