Đáp án đúng: B $\sqrt{{{n}^{2}}+{{a}^{2}}n}-\sqrt{{{n}^{2}}+\left( a+2 \right)n+1}\sim \sqrt{{{n}^{2}}}-\sqrt{{{n}^{2}}}=0\xrightarrow[{}]{}$ nhân lượng liên hợp: Ta có$\lim \left( \sqrt{{{n}^{2}}+{{a}^{2}}n}-\sqrt{{{n}^{2}}+\left( a+2 \right)n+1} \right)=\lim \frac{\left( {{a}^{2}}-a-2 \right)n-1}{\sqrt{{{n}^{2}}+n}+\sqrt{{{n}^{2}}+1}}$ $=\lim \frac{{{a}^{2}}-a-2-\frac{1}{n}}{\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{1}{{{n}^{2}}}}}=\frac{{{a}^{2}}-a-2}{2}=0\Leftrightarrow \left[ \begin{array}{l}a=-1\\b=2\end{array} \right..$ Chọn B.