Giải thích các bước giải:
1) Ta có:
$\begin{array}{l}
I = \sqrt {{n^2} - 2n + 8} - n\\
= \dfrac{{{n^2} - 2n + 8 - {n^2}}}{{\sqrt {{n^2} - 2n + 8} + n}}\\
= \dfrac{{ - 2n + 8}}{{\sqrt {{n^2} - 2n + 8} + n}}
\end{array}$
2) Ta có:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 24x + a}}{{{x^2} - 4}} = - 5\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 24x + a = \left( {x - 2} \right)\left( {x - \dfrac{a}{2}} \right)\\
\mathop {\lim }\limits_{x \to 2} \dfrac{{x - \dfrac{a}{2}}}{{x + 2}} = - 5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 24x = - \left( {2 + \dfrac{a}{2}} \right)x\\
\dfrac{{2 - \dfrac{a}{2}}}{{2 + 2}} = - 5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2 + \dfrac{a}{2} = 24\\
2 - \dfrac{a}{2} = - 20
\end{array} \right.\\
\Leftrightarrow a = 44
\end{array}$
Vậy $a=44$
