Đáp án:

$I = 1$

Giải thích các bước giải:

 Ta có:

$\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 10}}{{x - 1}} = 5\\
 \Rightarrow \mathop {\lim }\limits_{x \to 1} \left( {f\left( x \right) - 10} \right) = 0\\
 \Rightarrow f\left( 1 \right) - 10 = 0\\
 \Rightarrow f\left( 1 \right) = 10
\end{array}$

Lại có;

$\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 10}}{{x - 1}} = 5\\
 \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 10}}{{\sqrt x  - 1}}.\mathop {\lim }\limits_{x \to 1} \dfrac{1}{\sqrt x  + 1}  = 5\\
 \Rightarrow \dfrac{1}{2}\mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 10}}{{\sqrt x  - 1}} = 5\\
 \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 10}}{{\sqrt x  - 1}} = 10
\end{array}$

Khi đó:

$\begin{array}{l}
I = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 10}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt {4f\left( x \right) + 9}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 10}}{{\sqrt x  - 1}}.\mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\sqrt {4f\left( x \right) + 9}  + 3}}\\
 = 10.\dfrac{1}{{\sqrt {4.10 + 9}  + 3}}\\
 = 1
\end{array}$

Vậy $I = 1$