t=(2^x-1)/(2^x+1)在(-无穷,0)上有解,求t取值范围

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t=(2^x-1)/(2^x+1)在(-无穷,0)上有解,求t取值范围t=(2^x-1)/(2^x+1)在(-无穷,0)上有解,求t取值范围t=(2^x-1)/(2^x+1)在(-无穷,0)上有解,求

t=(2^x-1)/(2^x+1)在(-无穷,0)上有解,求t取值范围
t=(2^x-1)/(2^x+1)在(-无穷,0)上有解,求t取值范围

t=(2^x-1)/(2^x+1)在(-无穷,0)上有解,求t取值范围
已知t=[(2^x)-1]/[(2^x)+1],求t的取值范围.
去分母得t[(2^x)+1]=(2^x)-1;即有(t-1)(2^x)=-t-1,故得2^x=-(t+1)/(t-1);
于是得x=log₂[-(t+1)/(t-1)]
由-(t+1)/(t-1)>0,得(t+1)/(t-1)

t=(2^x-1)/(2^x+1)
t=1-2/[1/2^|x|+1]
1<1/[1/2^|x|+1]<3/2
2<2/[1/2^|x|+1]<3
-3<-2/[1/2^|x|+1]<-2
-2<1-2/[1/2^|x|+1]<-1
∴ t∈(-2,-1)