Bài giải

${{\log }_{2}}x+{{\log }_{2}}(x-1)=1\Leftrightarrow {{\log }_{2}}x(x-1)=1$

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

$\Leftrightarrow \left\{ \begin{align}& x>1 \\ & x(x-1)=2 \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x>1 \\ & {{x}^{2}}-x-2=0 \\ \end{align} \right.$ $\Leftrightarrow \left\{ \begin{align}& x>1 \\ & x=-1\vee x=2 \\\end{align} \right.\Leftrightarrow x=2$

 Chọn đáp án C