Chọn: Đáp án B

Xét biểu thức: $A=\sqrt{\frac{{{\log }_{a+1}}\left( a+2 \right)}{{{\log }_{a}}\left( a+1 \right)}}=\sqrt{{{\log }_{a+1}}a.{{\log }_{a+1}}\left( a+2 \right)}$

Áp dung bất đẳng thức AM-BM ta có:

$\sqrt{{{\log }_{a+1}}a.{{\log }_{a+1}}\left( a+2 \right)}\le \frac{{{\log }_{a+1}}a.{{\log }_{a+1}}\left( a+2 \right)}{2}<\frac{{{\log }_{a+1}}a.\left( a+2 \right)}{2}<\frac{{{\log }_{a+1}}{{\left( a+1 \right)}^{2}}}{2}=1$

Từ đó suy ra: $A<1\Leftrightarrow \sqrt{\frac{{{\log }_{a+1}}\left( a+2 \right)}{{{\log }_{a}}\left( a+1 \right)}}<1\Leftrightarrow {{\log }_{a}}\left( a+1 \right)>{{\log }_{a+1}}\left( a+2 \right)$