已知函数f(x)=2^x-1/[2^（绝对值x）]

若x0,
2^x-1/2^x=2
(2^x)^2-2*2^x-1=0
2^x=1+√2或2^x=1-√2
又x>0使得2^x>1,所以2^x=1+√2
x=log(2)(1+√2)
2^t(2^(2t)-1/2^(2t))+m(2^t-1/2^t)≥0
(2^t)^4+m*(2^t)^2-m-1≥0
(2^2t)^2+m*(2^2t)-m-1≥0
t∈[1,2],2^2t∈[4,16]
g(x)=x^2+mx-m-1=0
(x+m+1)(x-1)=0
则-m-1-5