Cho $f(x)$ là đa thức thỏa mãn \(\mathop {\lim }...

C

- Hướng dẫn giải

$\mathop {\lim }\limits_{x \to 3} \frac{{f(x) - 15}}{{x - 3}} = 12 \Rightarrow f(3) = 15$

$T = \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt[3]{{5f(x) - 11}} - 4}}{{{x^2} - x - 6}} = \mathop {\lim }\limits_{x \to 3} \frac{{5(f(x) - 15)}}{{\left( {x - 3} \right)\left( {x + 2} \right)\left( {{{\left( {\sqrt[3]{{5f(x) - 11}}} \right)}^2} + 4\sqrt[3]{{5f(x) - 11}} + 16} \right)}}$

$\begin{array}{l}  = \mathop {\lim }\limits_{x \to 3} \frac{{5.12}}{{\left( {x + 2} \right)\left( {{{\left( {\sqrt[3]{{5f(x) - 11}}} \right)}^2} + 4\sqrt[3]{{5f(x) - 11}} + 16} \right)}}\\  = \frac{{12}}{{{{\left( {\sqrt[3]{{5f(3) - 11}}} \right)}^2} + 4\sqrt[3]{{5f(3) - 11}} + 16}} = \frac{1}{4} \end{array}$