Điều kiện : $x>-1$

${{\log }_{2}}{}^{2}\left( x+1 \right)-{{\log }_{2}}\left( {{x}^{2}}+2x+1 \right)-3>0\Leftrightarrow {{\log }_{2}}{}^{2}\left( x+1 \right)-2{{\log }_{2}}\left( x+1 \right)-3>0$

Đặt $t={{\log }_{2}}\left( x+1 \right)$ ta được : ${{t}^{2}}-2t-3=0\Leftrightarrow \left[ \begin{align}  & t<-1 \\  & t>3 \\ \end{align} \right.$

$\Rightarrow \left[ \begin{align}  & {{\log }_{2}}\left( x+1 \right)<-1 \\  & {{\log }_{2}}(x+1)>3 \\ \end{align} \right.$$\Leftrightarrow \left[ \begin{align}  & 08 \\ \end{align} \right.$$\Leftrightarrow \left[ \begin{align}  & -17 \\ \end{align} \right.$