ĐKXĐ : $\frac{1-x}{1+x}>0\Leftrightarrow x\in (-1,1)$Đặt $log_2\frac{1-x}{1+x}=m$ khi đó
$\frac{4+m^2}{2+m}>2\Leftrightarrow \begin{cases}2+m>0 \\ 4+m^2>4+2m \end{cases}\Leftrightarrow \begin{cases}m>-2 \\ m^2>2m \end{cases}\Leftrightarrow -2<m<0 \vee m>2$
TH1: $-2<m<0\Leftrightarrow \frac{1}{4}<\frac{1-x}{1+x}<1\Leftrightarrow \begin{cases}4-4x>1+x \\ 1+x>1-x \end{cases}\Leftrightarrow \begin{cases}x<\frac{3}{5} \\ x>0 \end{cases}$
TH2: $m>2\Leftrightarrow \frac{1-x}{1+x}>4\Leftrightarrow 1-x>4+4x\Leftrightarrow x<-\frac{3}{5}$
Vậy $x\in (0,\frac{3}{5}) \vee (-1,-\frac{3}{5})$