Bài giải

$\log _{2}^{2}(x+1)-6{{\log }_{2}}\sqrt{x+1}+2=0$

$\Leftrightarrow \log _{2}^{2}(x+1)-6.\frac{1}{2}.{{\log }_{2}}(x+1)+2=0$

$\Leftrightarrow \log _{2}^{2}(x+1)-3.{{\log }_{2}}(x+1)+2=0$

$\Leftrightarrow \left\{ \begin{align}& x+1>0 \\& {{\log }_{2}}(x+1)=1\vee {{\log }_{2}}(x+1)=2 \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x>-1 \\& x+1=2\vee x+1=4 \\\end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x>-1 \\& x=1\vee x=3 \\ \end{align} \right.$

 $\Leftrightarrow \left[ \begin{align}& x=1 \\ & x=3 \\ \end{align} \right.$

Chọn đáp án B.